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

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

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.

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

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