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Math Help - Investment, interest and profit

  1. #1
    Senior Member Mukilab's Avatar
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    Investment, interest and profit

    The formula A=P(1+\frac{R}{100})^{n}


    If I want to reverse this to have minus profit (depreciates) and minus income do I do A=P(1+\frac{-r}{100})^{n}???
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    The formula A=P(1+\frac{R}{100})^{t}


    If I want to reverse this to have minus profit (depreciates) and minus income do I do A=P(1+\frac{-r}{100})^{t}???
    You can but you can still use the first formula. It just so happens that R < 0

    For example if a car worth 14000 depreciates by 4% annually.

    P = 14000
    R = -4

    A = P \left(1-\frac{R}{100}\right)^t =  14000 \left(1-\frac{4}{100}\right)^t



    In the compound interest formula it's usually better to have time as an exponent (time is denoted by t) although n works fine so use whichever is easier
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  3. #3
    Senior Member Mukilab's Avatar
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    Thanks. Solved
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  4. #4
    Senior Member Mukilab's Avatar
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    Actually I'm confused by the wording

    The company bought a new truck
    Each year the value of the truck depreciates by 20%
    The value of the new truck can be multiplied by a single number to find its value at the end of 4 years.

    I did P=A(1-\frac{20}{100})^{4}=A(0.4096)

    and P=A(1-\frac{20}{100})^0=A(1)=A

    So I did \frac{1}{0.496}

    /Don't think it's correct :/
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    Actually I'm confused by the wording

    The company bought a new truck
    Each year the value of the truck depreciates by 20%
    The value of the new truck can be multiplied by a single number to find its value at the end of 4 years.

    I did P=A(1-\frac{20}{100})^{4}=A(0.4096)

    and P=A(1-\frac{20}{100})^0=A(1)=A

    So I did \frac{1}{0.496}

    /Don't think it's correct :/

    I believe the number is 0.4096. From what I can gather the question is asking what percentage of the original value the truck has after 4 years. Mathematically:

    P_4 = k\,P_0 where the constant k is the number in question.

    k = \frac{P_4}{P_0} = \frac{0.4096A}{A} = 0.4096
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  6. #6
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    I believe the number is 0.4096. From what I can gather the question is asking what percentage of the original value the truck has after 4 years. Mathematically:

    P_4 = k\,P_0 where the constant k is the number in question.

    k = \frac{P_4}{P_0} = \frac{0.4096A}{A} = 0.4096

    Oh I didn't need to do anything more ^^

    THanks again
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