Answer
$15.71\%$
Work Step by Step
Step 1. Identify the given quantities: Loan $A_p=12500$, monthly payment $R=420$, number of years $3$, interest is compounded monthly.
Step 2. Assume the interest is $r$, we have the per period interest as $i=t/12$, total number of payments $n=3\times12=36$, and we can set up the equation as:
$R=\frac{iA_p}{1-(1+i)^{-n}}$ or $420=\frac{12500r}{12(1-(1+r/12)^{-36})}$ which leads to
$2.48r-1+(1+r/12)^{-36}=0$
Step 3. Graph the function obtained above as shown in the figure, we can find a zero at $(0.1571,0)$, thus the interest rate is $r=15.71\%$
