Question

Joel borrowed $18,000 for 6 years at simple interest to purchase a vehicle. If Joel repaid a total of $20,953.80, at what rate did he borrow the money?

Enter the percent with the percent symbol. If the percent is not a whole value, enter it as a decimal where the last digit is not zero and there is a zero before the decimal point for values less than 1. For example, if the answer is .35%, 0.35% should be entered. Round the percent to the nearest thousandth of a percent if needed.

Answer 1

`~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$20953.80\\ P=\textit{original amount deposited}\dotfill & \$18000\\ r=rate\to r\%\to (r)/(100)\\ t=years\dotfill &6 \end{cases} \\\\\\ 20953.80=18000[1+((r)/(100))(6)]\implies \cfrac{20953.80}{18000}=1+\cfrac{6r}{100}`

`\cfrac{20953.80}{18000}=\cfrac{100+6r}{100}\implies \cfrac{2095380}{18000}=100+6r\implies \cfrac{11641}{100}=100+6r \\\\\\ \cfrac{11641}{100}-100=6r\implies \cfrac{1641}{100}=6r\implies \cfrac{1641}{600}=r \\\\\\ \cfrac{547}{200}=r\implies \stackrel{\%}{2.735} = r`