Posted by liz c on Saturday, November 13, 2010 at 9:46pm.
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.)
- math - Reiny, Saturday, November 13, 2010 at 9:53pm
The instructions in your questions tell you how to do it.
Give it a try.
- math - liz c , Saturday, November 13, 2010 at 10:16pm
- math - liz c , Saturday, November 13, 2010 at 10:18pm
- math - Reiny, Saturday, November 13, 2010 at 10:28pm
1000(1.13)^2 = 1276.90
You are right.
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 , Saturday, November 13, 2010 at 10:40pm
I multipied first part times .02 and subtracted that twice
- math - liz c , Saturday, November 13, 2010 at 10:41pm
wheres the .98 from?
- math - liz c , Saturday, November 13, 2010 at 10:47pm
the answer was wrong
- math - Reiny, Saturday, November 13, 2010 at 10:48pm
Just like when money increases by 1+i,
when the money decreases you would take 1-i
in this case i = .02
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, Saturday, November 13, 2010 at 10:52pm
you said you multiplied your answer by 2% and then subtracted it twice ...
let's try that
2% of 1276.90 = 25.54
2% of 1251.36 = 25.03
subtracting that leaves 1226.33
Well, well, what do you say now?
- math - liz c , Saturday, November 13, 2010 at 10:56pm
i put that answe ina and it was wrong
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