An escalator is descending at constant speed. A walks down and takes
50 steps to reach the bottom. B runs down and takes 90 steps in the
same time as A takes 10 steps. How many steps are visible when the
escalator is not operating?
2006-07-14 06:00:39 · 5 answers · asked by yrk_86 1 in Science & Mathematics ➔ Mathematics
Impossible to answer without more information. All you know from this is that B runs 9 times faster than A walks - assuming the escalator moving when B is running.
[EDIT]
I'd go with Philo on this, but the question doesn't state that B runs all the way down in 90 steps. It just says B takes 90 steps in the same TIME it takes A to take 10. B could have taken 100 steps to reach the bottom.
gimb1960 is wrong too - it's got to be more than 50. As stated the problem is incompletely specified.
2006-07-14 06:51:18 · answer #1 · answered by Will 6 · 2⤊ 1⤋
2 people, 2 equations.
A takes 50 steps, and during that time the escalator moves N steps, and that traverses the distance, D. So
50 + N = D
B takes 90 steps, and during that time A has taken 1/5 of his 50 steps and the escalator has moved 1/5 of its N steps, so
90 + (1/5)N = D
50 + N = D
450 + N = 5D
400 = 4D
100 = D
The distance down the escalator is 100 steps.
2006-07-14 13:30:39 · answer #2 · answered by Philo 7 · 0⤊ 0⤋
50 - (2*50 - 90)/10 = 42
2006-07-14 08:02:12 · answer #3 · answered by gjmb1960 7 · 0⤊ 0⤋
None. When it's not operating the store is closed.
2006-07-14 06:30:45 · answer #4 · answered by Raymond 6 · 0⤊ 0⤋
Geeezzzzzzzz........what is this a test?
2006-07-14 06:07:48 · answer #5 · answered by ? 3 · 0⤊ 0⤋