Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 8 - Personal Finance - 8.6 Cars - Exercise Set 8.6 - Page 548: 32

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: \[\begin{align} & P=\frac{PMT\left[ 1-{{\left( 1+\frac{r}{n} \right)}^{-nt}} \right]}{\left( \frac{r}{n} \right)} \\ & =\frac{\$399\left[1-{{\left(1+\frac{0.09}{12}\right)}^{-12\times2}}\right]}{\left(\frac{0.09}{12}\right)}\\&=\frac{\$399\left[1-{{\left(1+0.0075\right)}^{-24}}\right]}{\left(0.0075\right)}\\&=\frac{\$399\left[1-{{\left(1.0075\right)}^{-24}}\right]}{\left(0.0075\right)}\end{align}\] \[\begin{align} & =\frac{\$399\left[1-{{\left(1.0075\right)}^{-24}}\right]}{\left(0.0075\right)}\\&=\frac{\$399\left[1-0.8358\right]}{\left(0.0075\right)}\\&=\frac{\$399\left[0.1642\right]}{\left(0.0075\right)}\\&=\frac{65.5158}{0.0075}\\\end{align}\] \[\begin{align} & =\$8,735.44\\&\simeq\$8,735\\\end{align}\]
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