Two women and two men entered a cafeteria and each pulled a ticket (There's a illustration, but I can't post it here... It's a rectangle with numbers from 5-100 that go by 5 cents, 10 cents, 15 cents...etc)
Anyway...

1. The four ordered the same food and had he same numeral punched out on their ticket (the retangle is the ticket).

2. Each of the four had exactly four coins.

3. The two women had the same amount of money in coins, though no denomination (value) of coin was held by both women; the two men had the same amount of money in coins, though no denominationof coin was held by both men.

4. Each of the four paid the exact amount inndicated by the numberal that was punched out on his or her ticket.

Which numeral was punched out on each ticket? NOTE: "Coins" may be pennies (1 cent), nickles (5 cents), dimes (10 cents), quarters (25 cents), half dollar (50 cents) or silver dollar.

(In reply to a solution by chakravarthi)

As the problem is actually stated, you are correct, but it is obvious that the problem intended that the two men have a different amount from the amount the two women had. Otherwise there is no unique solution. The price could equally well be 80 cents, or 40 cents or 30 cents. By showing that one couple had 80 cents (3 quarters, 1 nickel/1 half-dollar, 3 dimes) and the other couple had 40 cents (1 quarter, 3 nickels/4 dimes) the price must be 30 cents (1 quarter, 1 nickel/3 dimes).

Posted by TomM on 2003-01-10 19:13:34