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.
