Question

Cameron invests money in a bank account

which gathers compound interest each year.
After 4 years there is $736.80 in the account.
After 7 years there is $788.82 in the account.
Work out the annual interest rate of the bank
account.
Give your answer as a percentage to 1 d.p.

Answer 1

first off, let's check the equation for each year.

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$736.80\\ P=\textit{original amount deposited}\\ r=rate\to r\%\to (r)/(100)\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases} \\\\\\ 736.80=P\left(1+((r)/(100))/(1)\right)^(1\cdot 4)\implies 736.80=P\left(1+(r)/(100) \right)^4 \\\\[-0.35em] ~\dotfill`

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$788.82\\ P=\textit{original amount deposited}\\ r=rate\to r\%\to (r)/(100)\\ n= \begin{array}{llll} \textit{times it compounds per year} \end{array}\dotfill &1\\ t=years\dotfill &7 \end{cases} \\\\\\ 788.82=P\left(1+((r)/(100))/(1)\right)^(1\cdot 7)\implies 788.82=P\left(1+(r)/(100) \right)^7 \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{736.80=P\left(1+(r)/(100) \right)^4}\implies \cfrac{736.80}{\left(1+(r)/(100) \right)^4}=P \\\\\\ \stackrel{\textit{using the 2nd equation}}{788.82=P\left(1+(r)/(100) \right)^7}\implies \stackrel{\textit{substituting on the 2nd equation}}{788.82=\cfrac{736.80}{\left(1+(r)/(100) \right)^4}\left(1+(r)/(100) \right)^7}`

`788.82=736.80\cfrac{ ~~ \left(1+(r)/(100) \right)^7 ~~ }{\left(1+(r)/(100) \right)^4}\implies 788.82=736.80\left(1+(r)/(100) \right)^3 \\\\\\ \cfrac{788.82}{736.80}=\left(1+(r)/(100) \right)^3\implies \sqrt[3]{\cfrac{788.82}{736.80}}=1+\cfrac{r}{100}\implies \sqrt[3]{\cfrac{788.82}{736.80}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[3]{\cfrac{788.82}{736.80}}=100+r\implies 100\sqrt[3]{\cfrac{788.82}{736.80}}-100=r\implies {\Large \begin{array}{llll} \stackrel{\%}{2.3}\approx r \end{array}}`