Answer:
Step-by-step explanation:
1. The monthly rate is 1/12 of the annual rate, so r = 0.075/12 = 0.00625. The number of months is 12 times the number of years, so n = 6·12 = 72. Put these numbers into the given formula and evaluate it.
A = 400·(1.00625^72 -1)/(0.00625·1.00625^72) . . . . note we added parentheses to your given formula to define the denominator properly
= 400·(1.56611743 -1)/(0.00625·1.56611743)
= 226.446972/(0.00625·1.56611743)
= 226.446972/0.0097882339
= 23,134.61
Courtney can afford a car loan of up to $23,134.61.
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2. Though loans are available for up to 72 months, Tyresa wants to pay on a loan for only 4 years, or 48 months. The APRs for new and used cars translate to monthly rates of ...
new: 0.0279/12 ≈ 0.002325 = r
used: 0.0329/12 ≈ 0.0027416667 = r
The same formula can be used with ...
new: P = 350, r = 0.002325, n = 48 ⇒ A = 15879.04
used: P = 350, r = 0.0027416667, n = 48 ⇒ A = 15721.34
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Comment on formula evaluation
I find it much easier to do the calculation by writing a formula into a graphing calculator or spreadsheet. The tool can do the arithmetic much faster and there are fewer chances for mistakes.
It also helps to have the right formula to begin with. If you use the formulas written here, you will get wrong answers. The product of factors to the right of the division symbol (/) needs to be in parentheses. For the second problem, the formula should be the same as for the first problem. (We can't figure where (1+r)^2 came from.)
The principal loan amount is given making it the subject of the fixed
monthly payment formula.
Response (approximate values):
First part:
Second part:
The fixed monthly payment formula is given as follows;
[tex]M = \mathbf{\dfrac{P \cdot \dfrac{r}{12} \cdot \left(1 + \dfrac{r}{12} \right)^n }{ \left(1 + \dfrac{r}{12} \right)^n - 1}}[/tex]
Where;
M = Monthly payment = $400
P = The loan amount
r = The annual interest rate = 7.5% = 0.075
n = Number of periods = Number of months in 6 years = 6 × 12 = 72
Which gives;
[tex]P = \mathbf{\dfrac{M \cdot \left( \left(1 + \dfrac{r}{12} \right)^n - 1 \right)}{\dfrac{r}{12} \cdot \left(1 + \dfrac{r}{12} \right)^n}}[/tex]
Therefore;
[tex]P = \dfrac{400 \times \left( \left(1 + \dfrac{0.075}{12} \right)^{72} - 1 \right)}{\dfrac{0.075}{12} \cdot \left(1 + \dfrac{0.075}{12} \right)^{72}} \approx \mathbf{23,134.61}[/tex]
The maximum car loan Courtney can afford is approximately $23,134.61
Second question;
APR for new car = 2.79%
APR for used car = 3.29%
Number of months = 72
Maximum amount Tyresa wants to spend = $350
Therefore;
[tex]Maximum \ for\ P_{new} = \dfrac{350\times \left( \left(1 + \dfrac{0.0279}{12} \right)^{72} - 1 \right)}{\dfrac{0.0279}{12} \cdot \left(1 + \dfrac{0.0279}{12} \right)^{72}} \approx \mathbf{23,178.95}[/tex]
Maximum loan amount can Tyresa take for a new car is therefore;
The affordable loan amount for a used car, [tex]P_{used}[/tex], is found as follows;
[tex]Maximum \ for\ P_{used} = \dfrac{350\times \left( \left(1 + \dfrac{0.0279}{12} \right)^{72} - 1 \right)}{\dfrac{0.0279}{12} \cdot \left(1 + \dfrac{0.0279}{12} \right)^{72}} \approx 22,840.34[/tex]
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