Alan writes the equation A=500(1.25)t to figure out how much it will cost him for a one-year loan of $500 with an interest rate of 25% compounded only once. He wants to determine the annual percentage rate (APR) for a loan that would cost the same amount overall, if it is compounded monthly, instead of only once. Which equation should he use, and what is the APR?

A≈500(1.25)12tAPR≈1.88%

A≈500(1.0188)12tAPR≈22.56%

A≈500(1.0188)12tAPR≈1.88%

A≈500(1.2512)112tAPR≈14.55%

Respuesta :

Answer:

[tex]A=500(1.0188)^{12t}[/tex] , APR ≅ 22.56% 2nd answer

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

∵ The equation of one-year loan of $500 with an interest rate

   of 25% compounded only once is [tex]A=500(1.25)^{t}[/tex]

∴ t = 1

∴ A = 500(1.25) = 625

∵ A loan that would cost the same amount overall, if it is

    compounded monthly, instead of only once

∴ n = 12 ⇒ compounded monthly

∵ t = 1 and n = 12

∵ P = 500

- Use the formula of the compounded interest above

∴ [tex]A=500(1+\frac{r}{12})^{12}[/tex]

- Equate A by 625 (the value of money of the 1st equation)

∵ [tex]500(1+\frac{r}{12})^{12}=625[/tex]

- Divide both sides by 500

∴ [tex](1+\frac{r}{12})^{12}=1.25[/tex]

- Reverse the exponent and take it the the other side

∴ [tex]1+\frac{r}{12}=(1.25)^{\frac{1}{12}}[/tex]

- Subtract 1 from both sides

∴ [tex]\frac{r}{12}=(1.25)^{\frac{1}{12}}-1[/tex]

∴ [tex]\frac{r}{12}=0.01876926512[/tex]

- Multiply both sides by 12

∴ r ≅ 0.2256

- Multiply it by 100%

∴ r = 22.56%

The annual percentage rate (APR) is 22.56%

Substitute the value of r in the equation above

∴ [tex]A=500(1+\frac{0.2256}{12})^{12t}[/tex]

∵ [tex](1+\frac{0.2256}{12})=1.0188[/tex]

∴ [tex]A=500(1.0188)^{12t}[/tex]

He should use the equation [tex]A=500(1.0188)^{12t}[/tex]