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+(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* P \\ A=2*\text{ \$3500} \\ A=\text{ \$7000} \end{gathered}`

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

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

Hence,

The interest rate to the nearest tenth= 0.1