Reply To: The Riddle Thread….

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#1068010

Joseph-

Margarita: 6 x 1.5 = 9

Vodka: 5.75 x 1 = 5.75

Rum: 5 x 1 = 5

Pepsi: 2 x 1.5 = 3

Lemonade: 2 x 1 = 2

Water: 0 x 3 = 0

2) the only number that give six or more bills in change from $40 are 1,2,6,11 and 21. if margarita$ = a and lemonade$=b and a / 2 >= b (cond. 5) a can only be 1 or 2, b can only be 6 or 11. since a x 1.5 + b must be 11 or 21, the only combo that works is a (margarita) = 6, b (lemonade) = 2

3) water <= lemonade x 3. since only vodka has a non-integer price, water = 0.

4) The coin in change can only be a quarter (the whole dollar amounts of drinks can only be divided by 2, and vodka x 4 = an integer).

5) The total is over $20 (11 margarita and lemonade, at least 4 for the rum, at least 4.25 for the vodka an at least 1 for the pepsi). This means the change can either be 2.25, 6.25, 11.25 or 15.25 (two bills, one quarter).

7) Rum is 4,5 or 6 (floor = 2 x lemonade, ceiling = margarita). A second rum returns >= (six coins + bills) change. This only works if the second rum is 2.50. Rum = 5.

I used all the clues, so I don’t know which is extra.