# math

posted by 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 - ,

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

• math - ,

first part?1276.9

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• 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

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wheres the .98 from?

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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

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