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0 / 60 seg.
I have two coins, one is marked George I and one is marked George IV. One is genuine but one is a forgery. Which is the forgery?
G
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I
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A
C
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N
W
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L
D
N
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M
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K
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D
G
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G
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B
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S
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T
H
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M
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P
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D
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C
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D
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W
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L
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N
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H
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V
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B
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K
N
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G
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Clue
GEORGE I. A COIN WOULD NOT BE MARKED GOERGE I BECAUSE AT THE TIME IT WAS PRODUCED IT WOULD NOT HAVE BEEN KNOWN THAT THERE WAS GOING TO BE A GEORGE II
60 MINUTES. MANY OF YOU PROBABLY THINK THAT THEY SHOULD MULTIPLY 3 PILLS BY 30 MINUTES TO GIVE 90 MINUTES. BUT THINK OF HOW YOU TAKE THE PILLS. YOU TAKE THE FIRST PILL, AND IN A HALF-HOUR YOU TAKE THE SECOND PILL, AND THEN IN THE NEXT HALF-HOUR YOU TAKE THE THIRD PILL. SO THE NUMBER OF MINUTES THAT PASS FROM THE FIRST TO THE THIRD PILL IS 60 MINUTES
FIRST YOU SELECT A FRUIT FROM THE BOX MARKED PEACHES AND ORANGES. IF IT WAS A ORANGE YOU SELECTED, YOU KNOW THAT THE BOX COULD ONLY CONTAIN ORANGES. IF IT WAS A PEACH, YOU KNOW THAT THE BOX COULD ONLY CONTAIN PEACHES SINCE EACH BOX IS INCORRECTLY MARKED. IF, FOR EXAMPLE AN ORANGE WAS SELECTED, YOU WOULD MARK THAT BOX ORANGES AND SWITCH THE OTHER TWO INCORRECT LABELS AROUND. NOW ALL THREE WOULD BE CORRECTLY LABELED
THERE ARE 2 MEN, 5 WOMEN AND 13 CHILDREN. LET THE NUMBER OF MEN, WOMEN AND CHILDREN BE M, W, AND C RESPECTIVELY. THEN THE PROBLEM STATES THAT: M + W + C = 20 3M + 1.5W + 0.5C = 20 MULTIPLYING EQUATION 2 BY 2 GIVES: 6M + 3W + C = 40 SUBTRACTING THE ABOVE EQUATION FROM THE FIRST EQUATION GIVES: 5M + 2W = 20 THE UNIQUE SOLUTION IS M = 2, W = 5 AND C = 13
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