Easton has two options for buying a car. Option A is 2.1%2.1% APR financing over 3636 months and Option B is 6.1%6.1% APR over 3636 months with $2000$⁢2000 cash back, which he would use as part of the down payment. The price of the car is $25,083$⁢25,083 and Easton has saved $2500$⁢2500 for the down payment. Find the total amount Easton will spend on the car for each option if he plans to make monthly payments. Round your answers to the nearest cent, if necessary.

Let's break down the problem and compute the total amount Easton will spend on the car for each option.

### Step 1: Set up key information
- **Price of the car**: $25,083
- **Easton’s saved down payment**: $2,500
- **Loan duration**: 36 months (3 years)

### Option A:
- **APR (Annual Percentage Rate)**: 2.1%
- **Loan term**: 36 months
- **Easton’s total down payment**: $2,500
- **Loan amount** = Price of the car - Down payment  
  = $25,083 - $2,500 = $22,583

### Option B:
- **APR (Annual Percentage Rate)**: 6.1%
- **Loan term**: 36 months
- **Easton’s down payment**: $2,500 + $2,000 cash back = $4,500
- **Loan amount** = Price of the car - Down payment  
  = $25,083 - $4,500 = $20,583

### Step 2: Monthly Payment Formula
We can use the formula for monthly payments on an amortizing loan:

\[
M = \frac{P \cdot r(1 + r)^n}{(1 + r)^n - 1}
\]

Where:
- $ M $ = monthly payment
- $ P $ = loan amount (principal)
- $ r $ = monthly interest rate (APR divided by 12)
- $ n $ = total number of payments (loan term in months)

#### Option A:
- $ P = 22,583 $
- APR = 2.1%, so monthly rate $ r = \frac{2.1\%}{12} = \frac{0.021}{12} = 0.00175 $
- $ n = 36 $

#### Option B:
- $ P = 20,583 $
- APR = 6.1%, so monthly rate $ r = \frac{6.1\%}{12} = \frac{0.061}{12} = 0.005083 $
- $ n = 36 $

Let’s calculate the monthly payments for both options.Let's manually calculate the monthly payments and total cost.

### Option A:
- **Loan Amount**: $ P = 22,583 $
- **APR**: 2.1% → Monthly rate $ r = \frac{0.021}{12} = 0.00175 $
- **Loan Term**: $ n = 36 $ months

Using the monthly payment formula:

\[
M_A = \frac{22,583 \times 0.00175 \times (1 + 0.00175)^{36}}{(1 + 0.00175)^{36} - 1}
\]

### Option B:
- **Loan Amount**: $ P = 20,583 $
- **APR**: 6.1% → Monthly rate $ r = \frac{0.061}{12} = 0.005083 $
- **Loan Term**: $ n = 36 $ months

Similarly:

\[
M_B = \frac{20,583 \times 0.005083 \times (1 + 0.005083)^{36}}{(1 + 0.005083)^{36} - 1}
\]

Let me resolve these calculations for you.Here are the results:

### Option A:
- **Monthly payment**: \$647.82
- **Total amount spent** over 36 months: \$23,321.58

### Option B:
- **Monthly payment**: \$627.11
- **Total amount spent** over 36 months: \$22,575.88

### Conclusion:
- Option A (2.1% APR) results in a higher total cost (\$23,321.58) but slightly higher monthly payments (\$647.82).
- Option B (6.1% APR with \$2,000 cash back) results in a lower total cost (\$22,575.88) and slightly lower monthly payments (\$627.11).

Would you like more details or clarification on any part?

### Relative Questions:
1. How is APR applied to car loans?
2. How does a lower down payment affect total loan costs?
3. What are the advantages of cash-back offers in financing?
4. How does loan term length impact monthly payments?
5. Can early repayment lower total interest paid?

### Tip:
Always compare the **total cost** of a loan, not just the monthly payments, to fully assess its affordability.

Easton has two options for buying a car. Option A is 2.1% APR financing over 36 months and Option B is 6.1% APR over 36 months with $2000 cash back, which he would use as part of the down payment. The price of the car is $25,083 and Easton has saved $2500 for the down payment. Find the total amount Easton will spend on the car for each option if he plans to make monthly payments. Round your answers to the nearest cent, if necessary.

Explore how to compare two car loan options with different APRs and down payments. Learn to calculate monthly payments and the total cost using amortization formulas. This detailed guide includes practical financial math suitable for Grades 10-12.

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