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elevator problem

Expert replies
by ddm » Wed Aug 27, 2008 2:59 pm
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Source: — Problem Solving |

by parallel_chase » Wed Aug 27, 2008 3:53 pm
Steve's rate = 57 floors/minute
Joyce's rate = 63 floors/minute

Steve is at 11th floor
Joyce is at 51st floor

distance between joyce and steve = 51-11 = 40 floors

increasing or decreasing rate = 57+63 =120 floors/minute

D/R = T
40/120 = 1/3 minute or 20 seconds.

In 20 seconds Joyce would have covered 63*20/60 = 21 floors
In 20 seconds Steve would have covered 57*20/60 = 19 floors


51-21 = 30 (Joyce is going down)
11+19 = 30 (Steve is going up)

Hence answer is 30.
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by raju232007 » Thu Aug 28, 2008 2:46 am
parallel chase has given a good explanation...Let me provide another way to solve this problem...

Steve's rate = 57 floors/minute
Joyce's rate = 63 floors/minute

total distance between joyce and steve = 51-11 = 40 floors

In this type of problem the total distance is the sum of invidual distances...therefore

57/60t+63/60t= 40
2t=40
t=20s

distance covered by joyce=63*20/60=21 floors
distance covered by steve=57*20/60=19 floors

steve is going up from the 11th floor ...so 19+11=30
joyce is coming down from 51st floor....so 51-21=30
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Re: elevator problem

by sudi760mba » Sat Mar 21, 2009 3:30 pm
ddm wrote:attachment
Steve is on 11th floor traveling up 57floors/min
Joyce is on the 51st floor traveling down 63 floors/min

This is essentially a rate problem and if two objects are going towards each other, we would combine the rates.

Distance between floors: 51st - 11th = 40 floors to be covered between both of them
Rate at which they approach each other: 57 floors/min + 63 floors/min = 120 floors/min

40 floors to be covered / (120 floors/min) = 20 seconds


Steve in 20 seconds would be able to cover 19 floors (57 floors/min* 20 seconds) hence 19 + 11 = 30th floor

Joyce in 20 seconds would be able to cover 21 floors (63 floors/min * 20 seconds) 51 - 21 = 30th floor

They would meet on the 30th floor.
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