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

math 
Reiny,
The instructions in your questions tell you how to do it.
Give it a try. 
math 
liz c,
first part?1276.9

math 
liz c,
final answer?1225.82?

math 
Reiny,
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 
liz c,
I multipied first part times .02 and subtracted that twice

math 
liz c,
wheres the .98 from?

math 
liz c,
the answer was wrong

math 
Reiny,
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 
math 
Reiny,
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 
liz c,
i put that answe ina and it was wrong