Answer
\$8735
Work Step by Step
(a)
Compute the loan payment formula to prove unpaid balance formula shown as below:
\[PMT=\frac{\text{P}\times \left( \frac{r}{n} \right)}{\left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]}\]
Now, we have to multiply the \[\left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]\] in the above said formula.
\[PMT\times \left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]=\frac{\text{P}\times \left( \frac{r}{n} \right)}{\left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]}\times \left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]\]
Now the equation becomes as,
\[PMT\times \left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]=\text{P}\times \left( \frac{r}{n} \right)\]
Now divide both the sides with \[\left( \frac{r}{n} \right)\]we have,
\[\frac{PMT\times \left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]}{\left( \frac{r}{n} \right)}=\frac{\text{P}\times \left( \frac{r}{n} \right)}{\left( \frac{r}{n} \right)}\]
Now the equation results in as,
\[\frac{PMT\times \left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]}{\left( \frac{r}{n} \right)}=\text{P}\]
Re arranging the equation as follows:
\[P=\frac{PMT\times \left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]}{\left( \frac{r}{n} \right)}\]
(b)
Price of the car that is\[\$24,000\]. Down payment \[20%\]of the price of the car that is\[\frac{20}{100}\times \$24,000\,=\,\$4,800\]; Interest is\[9%\] for period of 5 years and \[n\,=\,12\]
Now, compute the adjusted Price of the car as shown below:
\[\begin{align}
& \text{Adjusted Price}=\text{Actual Price of the car}-\text{Down Payment} \\
& =\$24,000-\$4,800\\&=\$19,200\end{align}\]
Further, compute the amount of loan payment as shown below:
\[\begin{align}
& PMT=\frac{P\times \left( \frac{r}{n} \right)}{\left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]} \\
& =\frac{\$19,200\times\left(\frac{0.09}{12}\right)}{\left[1-{{\left(1+\frac{0.09}{12}\right)}^{-12\times5}}\right]}\\&=\frac{\$19,200\times\left(0.0075\right)}{\left[1-{{\left(1+0.0075\right)}^{-60}}\right]}\\&=\frac{\$19,200\times\left(0.0075\right)}{\left[1-{{\left(1.0075\right)}^{-60}}\right]}\\&\end{align}\]
\[\begin{align}
& =\frac{\$144}{\left[1-{{\left(1.0075\right)}^{-60}}\right]}\\&=\frac{\$144}{\left[1-{{\left(\frac{1}{1.0075}\right)}^{60}}\right]}\\&=\frac{\$144}{\left[1-{{\left(0.99255\right)}^{60}}\right]}\\&=\frac{\$144}{\left[1-0.63\right]}\\&\end{align}\]
\[\begin{align}
& =\frac{\$144}{0.37}\\&=\$389.1\\&\simeq\$399\end{align}\]
Also, compute the unpaid balance formula for three years as shown below:
\[\begin{align}
& P=\frac{PMT\left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]}{\left( \frac{r}{n} \right)} \\
& =\frac{\$399\left[1-{{\left(1+\frac{0.09}{12}\right)}^{-12\times2}}\right]}{\left(\frac{0.09}{12}\right)}\\&=\frac{\$399\left[1-{{\left(1+0.0075\right)}^{-24}}\right]}{\left(0.0075\right)}\\&=\frac{\$399\left[1-{{\left(1.0075\right)}^{-24}}\right]}{\left(0.0075\right)}\end{align}\]
\[\begin{align}
& =\frac{\$399\left[1-{{\left(1.0075\right)}^{-24}}\right]}{\left(0.0075\right)}\\&=\frac{\$399\left[1-0.8358\right]}{\left(0.0075\right)}\\&=\frac{\$399\left[0.1642\right]}{\left(0.0075\right)}\\&=\frac{65.5158}{0.0075}\\\end{align}\]
\[\begin{align}
& =\$8,735.44\\&\simeq\$8,735\\\end{align}\]
