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So in the end they each paid $9 dollars for the room
PhoenixOD
reply to post by CoherentlyConfused
This has been posted here before.
It not even a real maths problem as what is said is just not true
So in the end they each paid $9 dollars for the room
is not true so its a false premise.
edit on 6-3-2014 by PhoenixOD because: (no reason given)
PhoenixOD
reply to post by CoherentlyConfused
This has been posted here before.
It not even a real maths problem as what is said is just not true
So in the end they each paid $9 dollars for the room
is not true so its a false premise.
edit on 6-3-2014 by PhoenixOD because: (no reason given)
Actually, it's true. In the end each paid $9. The problem is that each was only charged $8-1/3 by the hotel. That means the bellboy took $2/3 from each guest. That's where he got his $2 from, 3 x 2/3 = 2.
Nope..they each paid 9.3333 recurring.
Dr. Rothman numbers the dollars:
Let's give each of the $30 a number from 1-30, keep track of each individual dollar, and see how the problem works.
Dollars numbered 1-30 are given to the manager. Then he wants to give $5 back, so he keeps the dollars numbered 1-25, and gives numbers 26-30 to the bellboy in the form of a five dollar bill. The bellboy splits up the five to get 5 one's: numbers 26, 27, 28, 29 and 30. He gives numbers 26, 27 and 28 to the customers and keeps numbers 29 and 30 for himself.
here is what an expert has to say
there was $30 and the bellboy took $2 which = 28/3 which is $9.3333333 each.
PhoenixOD
reply to post by DenyObfuscation
The expert says that dollars 1-25 and dollars 26,27,28 go to the people and 29, 30 go to the bellboy.
yet again 28/3 =9.3333333
If you cant account for all the dollars your method is flawed
edit on 6-3-2014 by PhoenixOD because: (no reason given)
PhoenixOD
reply to post by DenyObfuscation
The expert says that dollars 1-25 and dollars 26,27,28 go to the people and 29, 30 go to the bellboy.
yet again 28/3 =9.3333333
If you cant account for all the dollars your method is flawed
edit on 6-3-2014 by PhoenixOD because: (no reason given)