Question
Taylor wants to take out a mortgage of $380,000 with interest in these compounds monthly. Use the formula A= P (1 + r/n) n*t to find which of these loans will have the lowest total cost.
A. 20 years at 10% interest
B. 25 years at 6.5% interest
C. 30 years at 5% interest
D. 35 years at 4% interest
A. 20 years at 10% interest
B. 25 years at 6.5% interest
C. 30 years at 5% interest
D. 35 years at 4% interest
Answers
GPT-4o mini
To determine the total cost of each mortgage option, we will use the formula for compound interest given by:
\[ A = P \left(1 + \frac{r}{n}\right)^{n \cdot t} \]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = the principal amount (the initial amount of money, which is $380,000 in this case).
- \( r \) = annual interest rate (decimal).
- \( n \) = number of times that interest is compounded per year.
- \( t \) = the time the money is invested or borrowed for, in years.
For monthly compounding, \( n = 12 \).
Now, let's calculate the total cost for each option:
**A. 20 years at 10% interest:**
- \( P = 380,000 \)
- \( r = 0.10 \)
- \( n = 12 \)
- \( t = 20 \)
\[
A = 380000 \left(1 + \frac{0.10}{12}\right)^{12 \cdot 20}
\]
\[
A = 380000 \left(1 + 0.0083333 \right)^{240}
\]
\[
A = 380000 \left(1.0083333 \right)^{240}
\]
\[
A \approx 380000 \times 6.695
\]
\[
A \approx 2,544,100
\]
**B. 25 years at 6.5% interest:**
- \( r = 0.065 \)
- \( t = 25 \)
\[
A = 380000 \left(1 + \frac{0.065}{12}\right)^{12 \cdot 25}
\]
\[
A = 380000 \left(1 + 0.00541667 \right)^{300}
\]
\[
A = 380000 \left(1.00541667 \right)^{300}
\]
\[
A \approx 380000 \times 4.467
\]
\[
A \approx 1,696,600
\]
**C. 30 years at 5% interest:**
- \( r = 0.05 \)
- \( t = 30 \)
\[
A = 380000 \left(1 + \frac{0.05}{12}\right)^{12 \cdot 30}
\]
\[
A = 380000 \left(1 + 0.00416667 \right)^{360}
\]
\[
A = 380000 \left(1.00416667 \right)^{360}
\]
\[
A \approx 380000 \times 4.467
\]
\[
A \approx 1,696,600
\]
**D. 35 years at 4% interest:**
- \( r = 0.04 \)
- \( t = 35 \)
\[
A = 380000 \left(1 + \frac{0.04}{12}\right)^{12 \cdot 35}
\]
\[
A = 380000 \left(1 + 0.00333333 \right)^{420}
\]
\[
A = 380000 \left(1.00333333 \right)^{420}
\]
\[
A \approx 380000 \times 5.506
\]
\[
A \approx 2,088,800
\]
**Summary of Total Costs:**
- A (20 years at 10%): $2,544,100
- B (25 years at 6.5%): $1,696,600
- C (30 years at 5%): $1,696,600
- D (35 years at 4%): $2,088,800
The loans with the lowest total costs are options B and C, both approximately $1,696,600. Of these, since both are equal in total cost, you can choose either of them based on the duration.
\[ A = P \left(1 + \frac{r}{n}\right)^{n \cdot t} \]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = the principal amount (the initial amount of money, which is $380,000 in this case).
- \( r \) = annual interest rate (decimal).
- \( n \) = number of times that interest is compounded per year.
- \( t \) = the time the money is invested or borrowed for, in years.
For monthly compounding, \( n = 12 \).
Now, let's calculate the total cost for each option:
**A. 20 years at 10% interest:**
- \( P = 380,000 \)
- \( r = 0.10 \)
- \( n = 12 \)
- \( t = 20 \)
\[
A = 380000 \left(1 + \frac{0.10}{12}\right)^{12 \cdot 20}
\]
\[
A = 380000 \left(1 + 0.0083333 \right)^{240}
\]
\[
A = 380000 \left(1.0083333 \right)^{240}
\]
\[
A \approx 380000 \times 6.695
\]
\[
A \approx 2,544,100
\]
**B. 25 years at 6.5% interest:**
- \( r = 0.065 \)
- \( t = 25 \)
\[
A = 380000 \left(1 + \frac{0.065}{12}\right)^{12 \cdot 25}
\]
\[
A = 380000 \left(1 + 0.00541667 \right)^{300}
\]
\[
A = 380000 \left(1.00541667 \right)^{300}
\]
\[
A \approx 380000 \times 4.467
\]
\[
A \approx 1,696,600
\]
**C. 30 years at 5% interest:**
- \( r = 0.05 \)
- \( t = 30 \)
\[
A = 380000 \left(1 + \frac{0.05}{12}\right)^{12 \cdot 30}
\]
\[
A = 380000 \left(1 + 0.00416667 \right)^{360}
\]
\[
A = 380000 \left(1.00416667 \right)^{360}
\]
\[
A \approx 380000 \times 4.467
\]
\[
A \approx 1,696,600
\]
**D. 35 years at 4% interest:**
- \( r = 0.04 \)
- \( t = 35 \)
\[
A = 380000 \left(1 + \frac{0.04}{12}\right)^{12 \cdot 35}
\]
\[
A = 380000 \left(1 + 0.00333333 \right)^{420}
\]
\[
A = 380000 \left(1.00333333 \right)^{420}
\]
\[
A \approx 380000 \times 5.506
\]
\[
A \approx 2,088,800
\]
**Summary of Total Costs:**
- A (20 years at 10%): $2,544,100
- B (25 years at 6.5%): $1,696,600
- C (30 years at 5%): $1,696,600
- D (35 years at 4%): $2,088,800
The loans with the lowest total costs are options B and C, both approximately $1,696,600. Of these, since both are equal in total cost, you can choose either of them based on the duration.