Question

You invest $3,500 into an account where interest compounds monthly. If after 12 years your money has doubled, what was the interest rate? Round answers to the nearest tenth.

Answer 1

The formula for compound interest is given below as

$\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \text{where A=Amount} \\ P=\text{Principal} \\ r=\text{rate} \\ n=Number\text{ of times compounded(monthly therefore,n=12)} \\ t=\text{time in years=12 years} \end{gathered}$

Since after 12 years, the money doubled, this statement means that

$\begin{gathered} A=2\times P \\ A=2\times\text{ \ 3500} \\ A=\text{ \ 7000} \end{gathered}$

By substituting the values above in the formula, we will have

$\begin{gathered} \text{ \ 7000= \ 3500(1+}\frac{r}{12})^{12\times12} \\ \frac{\text{ \ 7000}}{3500}=(1+\frac{r}{12})^{144} \\ 2=(1+\frac{r}{12})^{144} \end{gathered}$$\begin{gathered} 2^{(\frac{1}{144})}=1+\frac{r}{12} \\ 1.00483=1+\frac{r}{12} \\ 1.00483-1=\frac{r}{12} \\ 0.00483=\frac{r}{12} \\ \text{cross multiply we will have} \\ r=0.00483\times12 \\ r=0.05796 \\ to\text{ the nearest tenth,} \\ r=0.1\text{ } \end{gathered}$

Hence,

The interest rate to the nearest tenth= 0.1