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?
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In the sleight of hand. More precisely, you're trying to add chicken to geese. – TZakrevskiy May 12 '14 at 17:29
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I know i just want to share interesting questions with everyone.@TZakrevskiy – Ravindra Sahay May 12 '14 at 17:30
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1How to drive a mathematician crazy. Part deux. IOW: Old hat. Ancient even. – Jyrki Lahtonen May 12 '14 at 17:31
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Well i never read that though cool one. I especially like these one way solution problems.@JyrkiLahtonen – Ravindra Sahay May 12 '14 at 17:33
2 Answers
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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.
Christian Blatter
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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.
$$\text{Thou shalt not think of cheating mathematics.}$$
Shaurya Gupta
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