Answer
(a) The monthly payments are $\$450$
The interest is $\$1200$
(b) The monthly payments are $\$293$
The interest is $\$2580$
(c) With Loan A, the monthly payments are $\$157$ more than the monthly payments with Loan B.
With Loan B, the interest is $\$1380$ 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) $PMT = \frac{P~(\frac{r}{n})}{[1-(1+\frac{r}{n})^{-nt}~]}$
$PMT = \frac{(\$15,000)~(\frac{0.051}{12})}{[1-(1+\frac{0.051}{12})^{-(12)(3)}~]}$
$PMT = \$450$
The monthly payments are $\$450$
We can find the total amount paid.
$\$450 \times 36 = \$16,200$
The interest is the difference between the total amount paid and the amount of the loan.
$I = \$16,200 - \$15,000 = \$1200$
The interest is $\$1200$
(b) $PMT = \frac{P~(\frac{r}{n})}{[1-(1+\frac{r}{n})^{-nt}~]}$
$PMT = \frac{(\$15,000)~(\frac{0.064}{12})}{[1-(1+\frac{0.064}{12})^{-(12)(5)}~]}$
$PMT = \$293$
The monthly payments are $\$293$
We can find the total amount paid.
$\$293 \times 60 = \$17,580$
The interest is the difference between the total amount paid and the amount of the loan.
$I = \$17,580 - \$15,000 = \$2580$
The interest is $\$2580$
(c) With Loan A, the monthly payments are $\$450$
With Loan B, the monthly payments are $\$293$
With Loan A, the monthly payments are $\$157$ more than the monthly payments with Loan B.
With Loan A, the interest is $\$1200$
With Loan B, the interest is $\$2580$
With Loan B, the interest is $\$1380$ more than the interest with Loan A.
