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: 4

Answer

(a) The monthly payments are $\$1219$ The interest is $\$3884$ (b) The monthly payments are $\$796$ The interest is $\$7760$ (c) With Loan A, the monthly payments are $\$423$ more than the monthly payments with Loan B. With Loan B, the interest is $\$3876$ more than the interest with Loan A.

Work Step by Step

We can use this formula to calculate the payments for a loan: $PMT = \frac{P~(\frac{r}{n})}{[1-(1+\frac{r}{n})^{-nt}~]}$ $PMT$ is the amount of the regular payment $P$ is the amount of the loan $r$ is the interest rate $n$ is the number of payments per year $t$ is the number of years (a) $PMT = \frac{P~(\frac{r}{n})}{[1-(1+\frac{r}{n})^{-nt}~]}$ $PMT = \frac{(\$40,000)~(\frac{0.061}{12})}{[1-(1+\frac{0.061}{12})^{-(12)(3)}~]}$ $PMT = \$1219$ The monthly payments are $\$1219$ We can find the total amount paid. $\$1219 \times 36 = \$43,884$ The interest is the difference between the total amount paid and the amount of the loan. $I = \$43,884 - \$40,000 = \$3884$ The interest is $\$3884$ (b) $PMT = \frac{P~(\frac{r}{n})}{[1-(1+\frac{r}{n})^{-nt}~]}$ $PMT = \frac{(\$40,000)~(\frac{0.072}{12})}{[1-(1+\frac{0.072}{12})^{-(12)(5)}~]}$ $PMT = \$796$ The monthly payments are $\$796$ We can find the total amount paid. $\$796 \times 60 = \$47,760$ The interest is the difference between the total amount paid and the amount of the loan. $I = \$47,760 - \$40,000 = \$7760$ The interest is $\$7760$ (c) With Loan A, the monthly payments are $\$1219$ With Loan B, the monthly payments are $\$796$ With Loan A, the monthly payments are $\$423$ more than the monthly payments with Loan B. With Loan A, the interest is $\$3884$ With Loan B, the interest is $\$7760$ With Loan B, the interest is $\$3876$ more than the interest with Loan A.
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