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Math Help - Interest Rate

  1. #1
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    Interest Rate

    Need help!

    At what interest rate, to the nearest hundredth of a percent, will $16,000 grow to $20,000 if invested for 5.25 years and interest rate compounded quarterly?
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  2. #2
    Junior Member Bradyns's Avatar
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    Re: Interest Rate

    IIRC

    C = P(1+\frac{p}{n})^n

    Where,
    C is the compounded amount
    P is the principle (starting) amount
    p is the rate (what you're after)
    n is time.

    Given that it is done quarterly for 5.25 years,
    n= (5.25)(4)
    n= 21

    C = 20,000
    P = 16,000

    Let's do some algebra with our n value as 21.

    Original Equation
    C = P(1+\frac{p}{21})^{(21)}

    Multiply out by P
    \frac{C}{P} = (1+\frac{p}{21})^{(21)}

    Take the 21st root of both sides.
    \sqrt[21]{\frac{C}{P}} = 1+\frac{p}{21}

    multiply both sides by 21, then subtract 21
    p = 21\left ( \sqrt[21]{\frac{C}{P}} \right )-21

    Plug in the values
    p = 21\left ( \sqrt[21]{\frac{20,000}{16,000}} \right )-21

    That should find you your rate.
    x = 21((1.25)^(1/21)) -21 - Wolfram|Alpha

    p = 22.43%
    Last edited by Bradyns; April 20th 2013 at 11:11 PM.
    Thanks from highlife2323
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  3. #3
    Junior Member Bradyns's Avatar
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    Re: Interest Rate

    Sorry, I got the equation wrong.

    This is the correct one:
    C = P\left ( 1 + \frac{p}{n} \right )^{nt}

    C is the compounded amount
    P is the principle (starting) amount
    p is the rate (what you're after)
    n is times compounded annually.
    t is the number of years.

    Doing the same sort of stuff..

    (20,000) = (16,000)\left ( 1 + \frac{p}{4} \right )^{21}

    \sqrt[21]{\frac{20,000}{16,000}} = 1 + \frac{p}{4}

    p = (4)\sqrt[21]{\frac{20,000}{16,000}}-(4)

    p = 0.0427302 or 4.3%

    x = 4((1.25)^(1/21)) -4 - Wolfram|Alpha
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