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Math Help - Continuous deposits

  1. #1
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    Continuous deposits

    a) Tiffany deposits $12 into an account at the end of each day in 1989 and 1990 and $15 at the end of each day in 1991. The account earns an annual effective interest rate of 9% in 1989 and 1990 and an annual effective interest rate of 12% in 1991. Find the amount in Tiffany’s account on December 31, 1991. (Answer: $16,502.58)

    b) Rework part (a) using the approximation that the deposits are made continuously. (Answer: $16,504.75)


    I think I'm pretty close, just missing something somewhere. I have
    12[(1.000236)^730 - 1 / .000236] + 15[(1.000311)^365 - 1 / .000311]
    = 15355.5
    (because (1+j)^4=1.045 and (1+j)^4=(1.025)^5)

    And also I'm not sure for part (b) how you would work it out differently if the deposits are continuous.

    Thank you!
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  2. #2
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    For part (a) you forgot to multiply the first part (The 2 years of $12 deposits) by 1.12 in order to bring it to the end of 1991, then you will get 16502.58
    ie.
    [12[(1.000236)^730 - 1 / .000236]](1.12) + 15[(1.000311)^365 - 1 / .000311] = 16502.58
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  3. #3
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    Ok awesome thank you!

    And for part (b), I think it's setup like:
    (12)(730)(1.12)(S bar angle 1) + (15)(365)(S bar angle 1) , where (S bar angle 1) = [(1+i)^1 - 1] / ln(1+i)

    but again I'm just slightly off...
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  4. #4
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    Quote Originally Posted by tbl9301 View Post
    I think I'm pretty close, just missing something somewhere. I have
    12[(1.000236)^730 - 1 / .000236] + 15[(1.000311)^365 - 1 / .000311]
    = 15355.5
    You forgot the interest on the balance at Dec.31/90 for year 1991. OK?
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