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Inflation is running 2% per year when you deposit $1000 in an account earning interest of 13% per year compounded annually. In constant dollars, how much money will you have two years from now? (Hint: First calculate the value of your account in two years' time, and then find its present value based on the inflation rate. Round your answer to the nearest cent.)
$ 1

  • math - ,

    The instructions in your questions tell you how to do it.
    Give it a try.

  • math - ,

    first part?1276.9

  • math - ,

    final answer?1225.82?

  • math - ,

    first part:
    1000(1.13)^2 = 1276.90

    You are right.

    Second part:
    1276.90(.98)^2 = 1226.33

    One step calculation:
    1000(.98)^2(1.13)^2 = 12236.33

    How did you get 1225.82 ?

  • math - ,

    I multipied first part times .02 and subtracted that twice

  • math - ,

    wheres the .98 from?

  • math - ,

    the answer was wrong

  • math - ,

    Just like when money increases by 1+i,
    when the money decreases you would take 1-i
    in this case i = .02
    so value(1-.02)^2

    when the rate was 13% did you not take 1 + .13 ?

    suppose we do it step by step

    start with 1000
    that loses 2% so amount left = .98(1000) = 980
    that gains 13%, so 980(1.13) = 1107.40

    that loses 2% , so amount left = 1107.4(.98) = 1085.25
    which gains 13% , 1085.25(1.13) = 1226.33

  • math - ,

    you said you multiplied your answer by 2% and then subtracted it twice ...

    let's try that

    2% of 1276.90 = 25.54
    leaving 1251.36

    2% of 1251.36 = 25.03
    subtracting that leaves 1226.33

    Well, well, what do you say now?

  • math - ,

    i put that answe ina and it was wrong

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