GRAPHIC MIND GAME ANSWERS!!!

Here are the answers for those graphic questions. I hope you enjoyed them. We will be puting up more graphic questions shortly.

  1. Number the coins from 1 to 12. Then weigh coins 1, 2, 3, and 4 against coins 5, 6, 7, and 8. If they balance weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin). If they balance, we know coin 12, the only unweighed one, is the counterfeit. The third weighing indicates whether it is heavy or light.
    If, however, at the second weighing (above), coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light. Weigh 9 and 10. If they balance, 11 is heavy. If they don't balance, either 9 or 10 is light.
    Now assume that at first weighing the side with coins 5, 6, 7, and 8 is heavier than the side with coins 1, 2, 3, and 4. This means that either 1, 2, 3, or 4 is light or 5, 6, 7, or 8 is heavy. Weigh 1, 2, and 5 against 3, 6, and 9. If they balance, it means that either 7 or 8 is heavy or 4 is light. By weighing 7 and 8 we obtain the answer, because if they balance then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit.
    If we weigh 1, 2, and 5 against 3, 6, and 9, the right side is heavier, then either 6 is heavy or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.
    If however, when we weigh 1, 2, and 5 against 3, 6, and 9, the right side is lighter, than either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.

  2. No. 1 and No. 3. Reason: No. 2 does not matter because the question is concerned only with red cards and if No. 4 is either red or green the answer could still be yes. If No. 1 has a circle the answer is no and if No. 3 is red the answer is no.

  3. Quite an easy one actually. Just remember that the question never made any mention of boundaries for the lines.

  4. Another practical lesson. By making symetrical cuts the image fits together as a square. (Neat-o.)

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