math
posted by liz c on .
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

The instructions in your questions tell you how to do it.
Give it a try. 
first part?1276.9

final answer?1225.82?

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 ? 
I multipied first part times .02 and subtracted that twice

wheres the .98 from?

the answer was wrong

Just like when money increases by 1+i,
when the money decreases you would take 1i
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 
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? 
i put that answe ina and it was wrong