Yes, and you answered that riddle right. But it's worded wrong, supposed to ask you this:


But it doesn't even make sense that way. When you calculate 9 X 3=27, that's already including the $2...they were supposed to pay $25, but he kept $2, so it's $27...plus the $3 that they already got back equals $30.

 

They paid $9 each...$25 that they owed for the room, plus $2 that the bellboy took equals $27/3=$9 even. They were suppose to pay ~$8.33 each...$25/3=$8.33.

 

they got $3 back :loco:

 

think outside the box.

Technically they paid $9 each cuz they gave a $10 and got $1 back. but mathmatically they paid $9.33 each.

The room is $25/3=$8.33each.
$5 is given back, only 3 is recieved.
So out of $30 they all paid $9 for the room.

but technically they paid $9.33

 

no, they paid $9 even. Think inside the box.

 

think outside the box?...
sorry but ur wrong

 

***k your boxes

lol

 

:naughty:

 

http://mathforum.org/library/drmath/view/57965.html

"
Date: 04/09/98 at 11:35:24
From: Doctor Rothman
Subject: Re: Where's the 30th dollar?

Hi. The first time I read it, I didn't get it either. But then I
realized that you just have to be REALLY careful about how you count
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.

-Doctor Rothman, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
"

we win

 

i actually knew that, but i couldn't word it. and i'm lazy.