Credit card A has an APR of 19.3% and an annual fee of $84, while credit card B has an APR of 24.6% and no annual fee. All else being equal, which of these equations can be used to solve for the principal P for which the cards offer the same deal over the course of a year? (Assume all interest is compounded monthly.)

The equation [tex]P(1+\frac{0.193}{12})^{12}+84=P(1+\frac{0.246}{12})^{12}[/tex] can be used to solve for the principal P for which the cards offer the same deal over the course of a year

Step-by-step explanation:

The formula of the compounded interest is [tex]A=P(1 + \frac{r}{n})^{nt}[/tex] , where

A is the future value of the investment/loan, including interest
P is the principal investment amount
r is the annual interest rate (decimal)
n is the number of times that interest is compounded per unit t
t is the time the money is invested or borrowed for

Card A:

∵ Credit card A has an APR of 19.3% and an annual fee of $84

∴ r = 19.3% = 19.3 ÷ 100 = 0.193

∴ Annual fee = 84

∵ Interest is compounded monthly for a year

∴ n = 12 and t = 1

- Substitute the values of r, n and t in the formula above

∴ [tex]A=P(1 + \frac{0.193}{12})^{12(1)}[/tex]

- Add the value of the annual fee

∴ [tex]A=P(1 + \frac{0.193}{12})^{12}+84[/tex]

Card B:

∵ Credit card B has an APR of 24.6% and no annual fee

∴ r = 24.6% = 24.6 ÷ 100 = 0.246

∴ Annual fee = 0

∵ Interest is compounded monthly for a year

∴ n = 12 and t = 1

- Substitute the values of r, n and t in the formula above

∴ [tex]A=P(1 + \frac{0.246}{12})^{12(1)}[/tex]

∴ [tex]A=P(1 + \frac{0.246}{12})^{12}[/tex]

- Equate The equations of cards A and B

∴ [tex]P(1+\frac{0.193}{12})^{12}+84=P(1+\frac{0.246}{12})^{12}[/tex]

The equation [tex]P(1+\frac{0.193}{12})^{12}+84=P(1+\frac{0.246}{12})^{12}[/tex] can be used to solve for the principal P for which the cards offer the same deal over the course of a year