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Solve the math problem

Thomas1016

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Mar 18, 2007
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3 friends check into a hotel the room rate is 30 dollars so each of the friends pays 10 dollars each. after they have gone up to thier room the manager realizes the room is actually 25 dollars and she has over charged the 3 friends so she sends a bellboy up to thier room with a 5 dollar refund. on the way up the bellboy realizes that he can not evenly split 5 dollars among 3 people so he decides to keep 2 dollars and give each of the friends 1 dollar each. The 3 friends recieve the the 3 dollar refund and are happy becuase now they are only in the room 9 dollars each.

The question is if the 3 friends are in the room 9 dollars each and 9x3=27 and the bellboy kept 2 dollars what happened to the other dollar?
 
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The 2 dollars the bellboy kept is already accounted for in the 27 dollars the friends paid. Adding it again to the amount the friends paid gives a meaningless sum.

Of the $30:
$25 is with the manager
$2 is with the bellboy
$1 is with friend 1
$1 is with friend 2
$1 is with friend 3
----
$30
 
3 friends check into a hotel the room rate is 30,000,000 dollars so each of the friends pays 10,000,000 dollars each. after they have gone up to thier room the manager realizes the room is actually 25 dollars and she has over charged the 3 friends so she sends a bellboy up to thier room with a 29,999,995 dollar refund. on the way up the bellboy realizes that he can not evenly split 29,999,995 dollars among 3 people so he decides to keep 2 dollars and give each of the friends 9,999,991 dollar each. The 3 friends recieve the the 9,999,991 dollar refund and are happy becuase now they are only in the room 9 dollars each.

The question is if the 3 friends are in the room 9 dollars each and 9x3=27 and the bellboy kept 2 dollars what happened to the other 29,999,971 dollars?
 
The Statue of Liberty was holding it when David Copperfield made her disappear.
 
quite correct! it is amazing how many people have a hard time seeing it though when they try to work it out in thier head.
 
I've never seen what the problem is. Any way you work it out the sums come out right.
 
3 guests (10 dollars/guests) = $30 dollars.

$30 = $25 dollars/manager * 1 manager + ($2/bellboy) * 1 bellboy + $3/ 3 guests *3 guests.

The Guests have tipped the bellboy $2/3 or 0.666666 each.
 
ShowerComic said:
3 guests (10 dollars/guests) = $30 dollars.

$30 = $25 dollars/manager * 1 manager + ($2/bellboy) * 1 bellboy + $3/ 3 guests *3 guests.

The Guests have tipped the bellboy $2/3 or 0.666666 each.
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So they were my fellow dutch countrymen huh?
 
I think that I first read this trick question in Mechanics Illustrated magazine 50 years ago. (when $30 was what a room cost)
 
It took me a little while to work out what happened. I didn't realize until after a few minutes that the prize to the room wasn't actually 27$ like the question made it seem but was 25$ as stated earlier. The question states that the room was actually 25$ and they paid 30$ so they should get a 5$ refund. each person gets 1$ meaning they paid 27$ all together and the bellboy gets 2$ so 1$ is missing. It makes it seem that the actual price of the room was 27$ when it was actually only 25$. Since it was only 25$ they actually paid 8.333..$ for the room and unwillingly tipped the bellboy 2$ (leaving 3 dollars which they kept).

8.33333x3=25+2=27+3=30.
 
Exactly. There's absolutely no confusion if you count down from the $30 and track who has what.

Manager has $30 and he gives the bellboy $5, leaving him with $25.
From the $5, the bellboy keeps $2 and gives each of the three guests $1.
$25 + $2 +$3 = $30.

By the way, to be nitpicky, this is arithmetic, not mathematics.
 
Big Al said:
Exactly. There's absolutely no confusion if you count down from the $30 and track who has what.

Manager has $30 and he gives the bellboy $5, leaving him with $25.
From the $5, the bellboy keeps $2 and gives each of the three guests $1.
$25 + $2 +$3 = $30.

By the way, to be nitpicky, this is arithmetic, not mathematics.
Click to expand...


Isn't arithmetic a field of mathematics?
 
Dustin Kesselberg said:
Isn't arithmetic a field of mathematics?
Click to expand...

No, it's a tool used in maths. Arithmetic is right down in the basic bedrock axioms of maths, like commutivity and associativity, multiplicative and adiditive identities (1 x n = n, 0 + n = n), etc. Algebra is where real maths starts.
 
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