Question

lsabella can afford a $410 per month car payment, and she's interested in either a sedan, which costs $21,600, or a station wagon, which costs $22,400. if she is being offered a 5 year car loan with an APR of 6% compounded monthly, which car can lsabella afford?

Answer 1

To determine which car Isabella can afford based on her budget of $410 per month, we can use the formula for the monthly payment on an amortizing loan:

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

Where:
- \( M \) = monthly payment
- \( P \) = loan principal (the amount borrowed)
- \( r \) = monthly interest rate (annual rate / 12)
- \( n \) = number of payments (loan term in months)

### Step 1: Calculate the monthly interest rate
The APR is 6%, so the monthly interest rate \( r \) can be calculated as:
\[
r = \frac{6\%}{12} = 0.005
\]

### Step 2: Calculate the number of payments
For a 5-year loan, the number of payments \( n \) is:
\[
n = 5 \times 12 = 60 \text{ months}
\]

### Step 3: Calculate the monthly payment for each car

**For the sedan ($21,600):**
\[
P = 21600
\]

Now substituting the values into the formula:
\[
M = 21600 \times \frac{0.005(1+0.005)^{60}}{(1+0.005)^{60} - 1}
\]

First, calculate \( (1 + 0.005)^{60} \):
\[
(1 + 0.005)^{60} \approx 1.34885 \quad (\text{using calculator})
\]
Thus,
\[
M = 21600 \times \frac{0.005 \times 1.34885}{1.34885 - 1}
\]
Calculating the numerator:
\[
0.005 \times 1.34885 \approx 0.00674425
\]
And the denominator:
\[
1.34885 - 1 = 0.34885
\]
Now the full term becomes:
\[
\frac{0.00674425}{0.34885} \approx 0.01933
\]
Therefore,
\[
M \approx 21600 \times 0.01933 \approx 417.88
\]

**For the station wagon ($22,400):**
\[
P = 22400
\]
Using the same formula:
\[
M = 22400 \times \frac{0.005(1+0.005)^{60}}{(1+0.005)^{60} - 1}
\]

Using the previously calculated \( (1 + 0.005)^{60} \):
\[
M = 22400 \times \frac{0.005 \times 1.34885}{0.34885}
\]
Calculating:
\[
0.005 \times 1.34885 \approx 0.00674425
\]
The monthly payment becomes:
\[
\frac{0.00674425}{0.34885} \approx 0.01933
\]
Therefore,
\[
M \approx 22400 \times 0.01933 \approx 433.91
\]

### Step 4: Compare to Isabella's budget
- **Sedan:** \( M \approx 417.88 \)
- **Station Wagon:** \( M \approx 433.91 \)

### Conclusion:
Isabella can afford the sedan as its monthly payment is approximately $417.88, which is just above her budget of $410. However, she cannot afford the station wagon since its payment of approximately $433.91 exceeds her budget. Thus, **Isabella can afford the sedan, but not the station wagon.**