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In 2010 a city raised $5,999,400 in tax revenue by charging each taxpayer the same amount. For 2011 there has been a net loss of 12 taxpayers, necessitating an increase fo $200 per taxpayer in order to maintain the tax revenue of $5,999,400. How many taxpayers were there in 2010?

  • math - ,

    Let n = number of taxpayers in 2010.
    So
    tax for each in 2010 = 5999400/n

    For 2011, there are 12 taxpayers less, and each one pays $200 more, so
    (n-12)*(5999400/n+200)=5999400
    Multiply each side by n:
    (n-12)(5999400+200n)=5999400
    Solve for n to get:
    -594 or 606. The negative root is discarded to leave 606 as the number of taxpayers in 2010.

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