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Math Help - Most Invested At 9 Percent

  1. #1
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    Most Invested At 9 Percent

    You are going to invest $30,000, part at 9% and part at 14%. What is the most that can be invested at 9% in order to make at least $3000 in interest per year?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    You can define a variable.

    Let x be the sum invested at 9% interest.
    (30000-x) will be the sum invested at 14% interest.

    The interest in the first case becomes: 9% \times x = \dfrac{9x}{100}

    The interest in the second case becomes: (30000-x) \times 14% = \dfrac{14(30000 - x)}{100}

    Total interest = \dfrac{9x}{100} + \dfrac{14(30000 - x)}{100}

    Now, equate to 3000 to find the value of x, the amount invested to 9% interest.

    \dfrac{9x}{100} + \dfrac{14(30000 - x)}{100} = 3000

    This gives you the greatest value invested at 9% interest to get 3000 interest per year.

    If less is invested at 9%, more will be invested at 14% and you get a higher interest.
    Ideally, if you want to make the most interest, put everything at 19%
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  3. #3
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    After doing the math on paper, I got x = -51,600. I got a negative value. Is this right? I don't think so.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    You should be getting $24000

    9x + 14(30000 - x) = 300000

    9x + 420000 - 14x = 300000

    -5x + 420000 = 300000

    Check your signs.
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  5. #5
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    -5x + 420000 = 300000

    -5x = 300000 - 420000

    -5x = -120000/-5

    x = 24,000

    I got it!
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