Symmetry wrote:Three men in a cafe order a meal the total cost of which is $15. They each contribute $5. The waiter takes the money to the chef who recognizes the three as friends and asks the waiter to return $5 to the men.
The waiter is not only poor at mathematics but dishonest and instead of going to the trouble of splitting the $5 between the three he simply gives them $1 each and pockets the remaining $2 for himself.
Now, each of the men effectively paid $4, the total paid is therefore $12. Add the $2 in the waiters pocket and this comes to $14.....where has the other $1 gone from the original $15?
As always, the answer is available online if you get stuck.
It didn't go anywhere.
The second part of the question is essentially rubbish and misleading.
now while the men only essentially paid 12 dollars, by giving the men back a dollar each the waiter effectively only lessened the cost of the meal to 12 instead of 10. this doesn't change the fact that the initial cost of the men was 5 (5x3=15), which was then reduced by 3 (5-2) to 12, and then the three dollars was distributed to each of the three men to make each of them get 1 dollar reducing their individual cost to 4 dollars. 4x3=12. If anything the dollar is "lost" because of the averaging occurring when distributing the money. This has to do with the fact that dividing and multiplying while reciprocal functions, if there is adding/subtracting (also reciprocal) will change the outcome of the function.
