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Math Help - double false position

  1. #1
    Junior Member
    Joined
    Sep 2008
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    double false position

    Hello everyone,

    Could someone please help me with this double false position problem?

    A buisness man doubled his money and he spent $12. He then left and doubled his money in another city and spent $12. He returned and doubled his money, and spent $12. At the end of the 3 trips and after the expenses he has $9 left. How much did he have at the begining?

    I know the answer is 11.625

    Do you see where I went wrong?

    the algebraic way:
    (2x-12)+(2x-12)+(2x-12)=9
    6x=45
    x=7.5

    Double false position:

    Guess $100

    ($100)2-12=188
    ($100)2-12=188
    ($100)2-12=188
    ________________
    564

    564-9=555

    Guess #2 $50

    (50)2-12=88
    (50)2-12=88
    (50)2-12=88
    __________
    264

    264-9=255


    100 555
    50 255
    cross multiply: (100)(255)=25500
    (50)(555)=27750

    Take the difference:
    27750-25500=2250
    2250/difference in error=
    2250/555-255 =
    7.5

    Thank you very much
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  2. #2
    Member
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    2(2(2x-12)-12)-12=9

    He's still the same man when he goes to another city. Plug the result of one city into the equation of the next city.
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  3. #3
    Junior Member
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    Thank you very much

    Could you please show me how to do it with double false position?

    Thank you
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  4. #4
    Junior Member
    Joined
    Sep 2008
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    Could someone please show me how to do this problem with double false position?

    Thank you
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