Question

A person invests $4000 at 2% interest compounded annually for 4 years and then invests the balance (the $4000 plus the interest earned) in an account at 8% interest for 7 years. Find the value of the investment after 11 years.

(Hint: You need to break this up into two steps/calculations. Be sure to round your balance at the end of the first 4 years to the nearest penny so you can use it in the second set of calculations.)

Answer 1

`\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$4000\\ r=rate\to 2\%\to (2)/(100)\to &0.02\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &4 \end{cases} \\\\\\ A=4000\left(1+(0.02)/(1)\right)^(1\cdot 4)\implies A=4000(1.02)^4\implies A\approx 4329.73`

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.


`\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$4329.73\\ r=rate\to 8\%\to (8)/(100)\to &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &7 \end{cases} \\\\\\ A=4329.73\left(1+(0.08)/(1)\right)^(1\cdot 7)\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396`

add both amounts, and that's her investment for the 11 years.