Three guys had a coffey in a diner. When they were done the waitress came so they could pay. It ended up to be 25 dollars. They suggested to split the bill three ways but they only had one 10 dollar bill each so they all gave the waitress one 10 dollar bill i.e. 30 dollars. She put the money in the register and came back with 5 dollar coins. Since you cant split 5 coins three ways the men took one dollar each and gave the waitress 2 dollars back as tap.
So the men gave her 10 dollars each and got one dollar back each meaning they had payed 9 dollars per person. The waitress got 2 dollars as tap.
9+9+9+2=29
Where did the 30th dollar go?
Math problem
Moderators: Duke, trewqh, korexus, Egbert
- Duke
- Moderator
- Posts: 1699
- Joined: Wed Sep 04, 2002 7:00 am
- Location: Sweden, Valn Ohtar
Math problem
First one here, last one to leave.
- Polymorphic
- Trooper
- Posts: 159
- Joined: Thu Jan 30, 2003 8:00 am
- Location: Oxfordish
- Contact:
Re: Math problem
25 for dinner, 3 each as change, 2 for waitress = 30, eh?Duke wrote:
9+9+9+2=29
Where did the 30th dollar go?
9+9+9-2=25
each man _pays_ 9, so you can't add the 2 - you subtract it.
- Duke
- Moderator
- Posts: 1699
- Joined: Wed Sep 04, 2002 7:00 am
- Location: Sweden, Valn Ohtar
I too see the logic here. 25 is in the register, 3 is with the customers and 2 is with the waitress.
The question here is why doesnt the math add up when I count the way I do.
If a person gave the waitress 10 bucks and get one back he has per definition paid 9 dollars.
If the waitress has gotten 2 dollars for tap she has two dollars.
So.....three times 9 dollars is 27 dollars. The waitress has two. I am still wondering where that 30ieth dollar is.
(I understand that it isnt really a mathproblem but agree that it is a tad wierd...oh and Poly, you are wierd to begin with so this is all perfectly clear to you )
The question here is why doesnt the math add up when I count the way I do.
If a person gave the waitress 10 bucks and get one back he has per definition paid 9 dollars.
If the waitress has gotten 2 dollars for tap she has two dollars.
So.....three times 9 dollars is 27 dollars. The waitress has two. I am still wondering where that 30ieth dollar is.
(I understand that it isnt really a mathproblem but agree that it is a tad wierd...oh and Poly, you are wierd to begin with so this is all perfectly clear to you )
First one here, last one to leave.
- korexus
- Moderator
- Posts: 2827
- Joined: Tue Nov 12, 2002 8:00 am
- Location: Reading
- Contact:
Because you're counting the 2 dollars the waitress has twice, it's included in the 27 dollars that the men have given out. There isn't actually a missing dollar, there are three. (You didn't count the three that the men keep at any point.) If you inlcude them, you'll get to 32 by your method so it's easy to see where the mistake is.Duke wrote:I too see the logic here. 25 is in the register, 3 is with the customers and 2 is with the waitress.
The question here is why doesnt the math add up when I count the way I do.
If a person gave the waitress 10 bucks and get one back he has per definition paid 9 dollars.
If the waitress has gotten 2 dollars for tap she has two dollars.
So.....three times 9 dollars is 27 dollars. The waitress has two. I am still wondering where that 30ieth dollar is.
I have a sneaky feeling that you know this and are just being contrary.
korexus.
With Great Power comes Great Irritability