Three friends problem.

Question

Once there were 3 friends: A, B, and C. They went together to have a lunch at a hotel. The lunch they had cost \$60 according to the menu. To pay the bill, each one contributed \$20. But when the waiter brought the sum to the manager, the manager returned \$10, saying today is a special day. But when the waiter went to return the money, he found that the three friends had left. So he got the member book, found the address, and left to return the money via taxi. He kept \$4 for his going and returning and returned \$2 to each of the friends. Now, each of the friends got \$2, so the money they contributed was $20-2= \$18$. Including the \$4 kept by the waiter, the total money is $(\$18*3) + \$4= \$58$. Where are the remaining \$2?

Answer 1

The $18\cdot3+4$ part is wrong. If The money each person initially spends is $20$. Then each person is supposed to get $3\frac{1}{3}$ back. However, the waiter takes away $\frac43$ per person for coming and going. So Each person therefore actually gets $\frac{10}3-\frac43=2$ back. In this process the total remains same.

$$\text{Thou shalt not think of cheating mathematics.}$$

Answer 2

This is indeed an old hat. Each of the three paid $\$\, 18.00$, makes $\$\,54.00$ in all. $\$\,4.00$ went to the waiter and $\$\,50.00$ to the manager. Nothing went lost.