Question
Emma wants to take a loan of $10,000 with interest that compound monthly. Use the formula A = P (1 + r/n) n*t to find which of these loan terms will have the lowest total cost.
A. 2 years at 3% interest
B. 3 years at 3% interest
C. 4 years at 3% interest
D. 5 years at 3% interest
A. 2 years at 3% interest
B. 3 years at 3% interest
C. 4 years at 3% interest
D. 5 years at 3% interest
Answers
GPT-4o mini
To determine which loan term has the lowest total cost, we can use the compound interest formula provided:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = the principal amount (the initial sum of money, in this case, $10,000).
- \( r \) = 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.
For this scenario, the principal \( P \) is $10,000, the interest rate \( r \) is 3% (or 0.03), and since the interest is compounded monthly, \( n \) is 12.
Let's calculate the total cost \( A \) for each loan option:
### A. 2 years at 3%
- \( t = 2 \)
\[
A = 10000 \left(1 + \frac{0.03}{12}\right)^{12 \times 2}
\]
\[
A = 10000 \left(1 + 0.0025\right)^{24}
\]
\[
A = 10000 \left(1.0025\right)^{24} \approx 10000 \times 1.061488 \approx 10614.88
\]
### B. 3 years at 3%
- \( t = 3 \)
\[
A = 10000 \left(1 + \frac{0.03}{12}\right)^{12 \times 3}
\]
\[
A = 10000 \left(1.0025\right)^{36} \approx 10000 \times 1.093443 \approx 10934.43
\]
### C. 4 years at 3%
- \( t = 4 \)
\[
A = 10000 \left(1 + \frac{0.03}{12}\right)^{12 \times 4}
\]
\[
A = 10000 \left(1.0025\right)^{48} \approx 10000 \times 1.126162 \approx 11261.62
\]
### D. 5 years at 3%
- \( t = 5 \)
\[
A = 10000 \left(1 + \frac{0.03}{12}\right)^{12 \times 5}
\]
\[
A = 10000 \left(1.0025\right)^{60} \approx 10000 \times 1.159274 \approx 11592.74
\]
### Summary of Total Amounts:
- A (2 years): approximately $10,614.88
- B (3 years): approximately $10,934.43
- C (4 years): approximately $11,261.62
- D (5 years): approximately $11,592.74
The loan term that has the lowest total cost is **A (2 years at 3% interest)** with a total payment of approximately **$10,614.88**.
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = the principal amount (the initial sum of money, in this case, $10,000).
- \( r \) = 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.
For this scenario, the principal \( P \) is $10,000, the interest rate \( r \) is 3% (or 0.03), and since the interest is compounded monthly, \( n \) is 12.
Let's calculate the total cost \( A \) for each loan option:
### A. 2 years at 3%
- \( t = 2 \)
\[
A = 10000 \left(1 + \frac{0.03}{12}\right)^{12 \times 2}
\]
\[
A = 10000 \left(1 + 0.0025\right)^{24}
\]
\[
A = 10000 \left(1.0025\right)^{24} \approx 10000 \times 1.061488 \approx 10614.88
\]
### B. 3 years at 3%
- \( t = 3 \)
\[
A = 10000 \left(1 + \frac{0.03}{12}\right)^{12 \times 3}
\]
\[
A = 10000 \left(1.0025\right)^{36} \approx 10000 \times 1.093443 \approx 10934.43
\]
### C. 4 years at 3%
- \( t = 4 \)
\[
A = 10000 \left(1 + \frac{0.03}{12}\right)^{12 \times 4}
\]
\[
A = 10000 \left(1.0025\right)^{48} \approx 10000 \times 1.126162 \approx 11261.62
\]
### D. 5 years at 3%
- \( t = 5 \)
\[
A = 10000 \left(1 + \frac{0.03}{12}\right)^{12 \times 5}
\]
\[
A = 10000 \left(1.0025\right)^{60} \approx 10000 \times 1.159274 \approx 11592.74
\]
### Summary of Total Amounts:
- A (2 years): approximately $10,614.88
- B (3 years): approximately $10,934.43
- C (4 years): approximately $11,261.62
- D (5 years): approximately $11,592.74
The loan term that has the lowest total cost is **A (2 years at 3% interest)** with a total payment of approximately **$10,614.88**.