Question

Mirta has a savings account which gathers compound interest each year. She opened the account with a deposit of 3000

. After 4 years there is 3690.37
in the account. Work out the annual interest rate of the savings account. Give your answer as a percentage to 1 d.p.

Answer 1

5.3 is about r

Explanation:

Answer 2

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$ 3690.37\\ P=\textit{original amount deposited}\dotfill &\$3000\\ r=rate\to r\%\to (r)/(100)\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{each year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}`

`3690.37 = 3000\left(1+((r)/(100))/(1)\right)^(1\cdot 4) \implies \cfrac{3690.37 }{3000}=\left(1+\cfrac{r}{100} \right)^4 \\\\\\ \sqrt[4]{\cfrac{3690.37 }{3000}}=1+\cfrac{r}{100}\implies \sqrt[4]{\cfrac{3690.37 }{3000}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[4]{\cfrac{3690.37 }{3000}}=100+r\implies 100\sqrt[4]{\cfrac{3690.37 }{3000}}-100=r\implies \boxed{5.3\approx r}`