Question
Aaron wants to take out a mortgage of $540,000 with interest that 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 7.5% interest
B) 25 years at 7% interest
C) 30 years at 4.5% interest
D) 35 years at 5.5% interest
A) 20 years at 7.5% interest
B) 25 years at 7% interest
C) 30 years at 4.5% interest
D) 35 years at 5.5% interest
Answers
GPT-4o mini
To determine which of the loans will have the lowest total cost, we can use the formula for compound interest:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \(A\) = the amount of money accumulated after n years, including interest.
- \(P\) = the principal amount (the initial loan amount).
- \(r\) = the annual interest rate (decimal).
- \(n\) = the number of times that interest is compounded per year.
- \(t\) = the number of years the money is borrowed for.
Given the principal \(P = 540,000\), we will calculate \(A\) for each option:
**Option A:** 20 years at 7.5% interest
- \(r = 7.5\% = 0.075\)
- \(n = 12\) (monthly compounding)
- \(t = 20\)
\[
A = 540,000 \left(1 + \frac{0.075}{12}\right)^{12 \cdot 20}
\]
\[
= 540,000 \left(1 + 0.00625\right)^{240}
\]
\[
= 540,000 \left(1.00625\right)^{240}
\]
Using a calculator, we find:
\[
A \approx 540,000 \cdot 4.392
\]
\[
A \approx 2,373,084
\]
**Option B:** 25 years at 7% interest
- \(r = 7\% = 0.07\)
- \(t = 25\)
\[
A = 540,000 \left(1 + \frac{0.07}{12}\right)^{12 \cdot 25}
\]
\[
= 540,000 \left(1 + 0.0058333\right)^{300}
\]
\[
= 540,000 \left(1.0058333\right)^{300}
\]
Using a calculator:
\[
A \approx 540,000 \cdot 5.352
\]
\[
A \approx 2,895,621
\]
**Option C:** 30 years at 4.5% interest
- \(r = 4.5\% = 0.045\)
- \(t = 30\)
\[
A = 540,000 \left(1 + \frac{0.045}{12}\right)^{12 \cdot 30}
\]
\[
= 540,000 \left(1 + 0.00375\right)^{360}
\]
\[
= 540,000 \left(1.00375\right)^{360}
\]
Using a calculator:
\[
A \approx 540,000 \cdot 3.849
\]
\[
A \approx 2,080,956
\]
**Option D:** 35 years at 5.5% interest
- \(r = 5.5\% = 0.055\)
- \(t = 35\)
\[
A = 540,000 \left(1 + \frac{0.055}{12}\right)^{12 \cdot 35}
\]
\[
= 540,000 \left(1 + 0.0045833\right)^{420}
\]
\[
= 540,000 \left(1.0045833\right)^{420}
\]
Using a calculator:
\[
A \approx 540,000 \cdot 6.438
\]
\[
A \approx 3,472,032
\]
**Summary of Total Costs:**
- A) $2,373,084
- B) $2,895,621
- C) $2,080,956
- D) $3,472,032
Thus, the loan that will have the lowest total cost is **Option C**: 30 years at 4.5% interest, with a total cost of approximately **$2,080,956**.
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \(A\) = the amount of money accumulated after n years, including interest.
- \(P\) = the principal amount (the initial loan amount).
- \(r\) = the annual interest rate (decimal).
- \(n\) = the number of times that interest is compounded per year.
- \(t\) = the number of years the money is borrowed for.
Given the principal \(P = 540,000\), we will calculate \(A\) for each option:
**Option A:** 20 years at 7.5% interest
- \(r = 7.5\% = 0.075\)
- \(n = 12\) (monthly compounding)
- \(t = 20\)
\[
A = 540,000 \left(1 + \frac{0.075}{12}\right)^{12 \cdot 20}
\]
\[
= 540,000 \left(1 + 0.00625\right)^{240}
\]
\[
= 540,000 \left(1.00625\right)^{240}
\]
Using a calculator, we find:
\[
A \approx 540,000 \cdot 4.392
\]
\[
A \approx 2,373,084
\]
**Option B:** 25 years at 7% interest
- \(r = 7\% = 0.07\)
- \(t = 25\)
\[
A = 540,000 \left(1 + \frac{0.07}{12}\right)^{12 \cdot 25}
\]
\[
= 540,000 \left(1 + 0.0058333\right)^{300}
\]
\[
= 540,000 \left(1.0058333\right)^{300}
\]
Using a calculator:
\[
A \approx 540,000 \cdot 5.352
\]
\[
A \approx 2,895,621
\]
**Option C:** 30 years at 4.5% interest
- \(r = 4.5\% = 0.045\)
- \(t = 30\)
\[
A = 540,000 \left(1 + \frac{0.045}{12}\right)^{12 \cdot 30}
\]
\[
= 540,000 \left(1 + 0.00375\right)^{360}
\]
\[
= 540,000 \left(1.00375\right)^{360}
\]
Using a calculator:
\[
A \approx 540,000 \cdot 3.849
\]
\[
A \approx 2,080,956
\]
**Option D:** 35 years at 5.5% interest
- \(r = 5.5\% = 0.055\)
- \(t = 35\)
\[
A = 540,000 \left(1 + \frac{0.055}{12}\right)^{12 \cdot 35}
\]
\[
= 540,000 \left(1 + 0.0045833\right)^{420}
\]
\[
= 540,000 \left(1.0045833\right)^{420}
\]
Using a calculator:
\[
A \approx 540,000 \cdot 6.438
\]
\[
A \approx 3,472,032
\]
**Summary of Total Costs:**
- A) $2,373,084
- B) $2,895,621
- C) $2,080,956
- D) $3,472,032
Thus, the loan that will have the lowest total cost is **Option C**: 30 years at 4.5% interest, with a total cost of approximately **$2,080,956**.