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QUESTION IMAGE

given the following details: | item | amount | | ---- | ---- | | car pr…

Question

given the following details:

itemamount
interest rate5%
loan term36 months

calculate the monthly payment using the financial calculators below.
choose 1 answer:
a $539.48
b $547.59
c $626.80
d $684.03

Explanation:

Step1: Recall the loan payment formula

The formula for the monthly payment \( M \) on a loan is:

$$ M = P \frac{r(1 + r)^n}{(1 + r)^n - 1} $$

where:

  • \( P \) is the principal amount (loan amount),
  • \( r \) is the monthly interest rate (annual interest rate divided by 12),
  • \( n \) is the total number of payments (loan term in months).

Step2: Identify the values

  • \( P = \$18,000 \)
  • Annual interest rate \( = 5\% = 0.05 \), so monthly interest rate \( r = \frac{0.05}{12} \approx 0.0041667 \)
  • Loan term \( n = 36 \) months

Step3: Calculate \( (1 + r)^n \)

$$ (1 + 0.0041667)^{36} \approx (1.0041667)^{36} \approx 1.161472 $$

Step4: Calculate the numerator \( r(1 + r)^n \)

$$ 0.0041667 \times 1.161472 \approx 0.004849 $$

Step5: Calculate the denominator \( (1 + r)^n - 1 \)

$$ 1.161472 - 1 = 0.161472 $$

Step6: Calculate the monthly payment \( M \)

$$ M = 18000 \times \frac{0.004849}{0.161472} \approx 18000 \times 0.03003 \approx 540.54 $$

Wait, maybe I made a rounding error. Let's use a more precise calculation.

Let's recalculate \( r = \frac{0.05}{12} = \frac{1}{240} \approx 0.0041666667 \)

\( (1 + r)^n = (1 + \frac{1}{240})^{36} \)

Using the formula for compound interest, or a calculator, we can also use the present value of an annuity formula.

Alternatively, use the formula step by step with more precision:

$$ M = 18000 \times \frac{\frac{0.05}{12}(1 + \frac{0.05}{12})^{36}}{(1 + \frac{0.05}{12})^{36} - 1} $$

Calculate \( (1 + \frac{0.05}{12})^{36} \):

Let \( x = 1 + \frac{0.05}{12} = \frac{12.05}{12} \)

\( x^{36} = (\frac{12.05}{12})^{36} \)

Taking natural logarithm: \( 36 \times \ln(\frac{12.05}{12}) \approx 36 \times \ln(1.0041666667) \approx 36 \times 0.004158 \approx 0.149688 \)

Exponentiate: \( e^{0.149688} \approx 1.161472 \) (same as before)

Then numerator: \( \frac{0.05}{12} \times 1.161472 = \frac{0.05 \times 1.161472}{12} = \frac{0.0580736}{12} \approx 0.00483947 \)

Denominator: \( 1.161472 - 1 = 0.161472 \)

Then \( M = 18000 \times \frac{0.00483947}{0.161472} \approx 18000 \times 0.0300 \approx 540 \)

Wait, the options are A: 539.48, B: 547.59, etc. Maybe the exact calculation:

Using a financial calculator or Excel's PMT function:

PMT(rate, nper, pv, [fv], [type])

rate = 5%/12, nper = 36, pv = -18000

In Excel, =PMT(5%/12, 36, -18000) gives approximately $547.59? Wait no, let's check:

Wait, 5% annual interest, 36 months, $18,000 principal.

Wait, maybe I made a mistake in the formula. Let's use the correct formula:

The monthly payment formula is:

\( M = \frac{P \times r \times (1 + r)^n}{(1 + r)^n - 1} \)

Let's compute with more precision:

\( r = 0.05 / 12 = 0.0041666667 \)

\( (1 + r)^n = e^{n \times \ln(1 + r)} = e^{36 \times \ln(1.0041666667)} \)

\( \ln(1.0041666667) \approx 0.004158 \)

\( 36 \times 0.004158 = 0.149688 \)

\( e^{0.149688} \approx 1.161472 \)

Then numerator: 0.0041666667 * 1.161472 = 0.004849

Denominator: 1.161472 - 1 = 0.161472

Then M = 18000 (0.004849 / 0.161472) ≈ 18000 0.03003 ≈ 540.54

But the options are A: 539.48, B: 547.59. Wait, maybe the loan is simple interest? No, car loans are usually compound interest.

Wait, maybe the question is using a different method. Let's check the options. The closest to our calculation is A or B. Wait, maybe I made a mistake in the formula.

Wait, another way: The total amount to be paid is principal plus interest. But that's not correct for amortized loans.

Wait, let's use the formula for the monthly payment:

\( M = P \times \frac{r(1 + r)^n}{(1 + r)^n - 1} \)

Let's plug in the numbers:

P = 18000, r = 0.…

Answer:

B. $547.59