Question
Jayden wants to take out a loan of $2,500 with interest that compounds 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. 3 years at 3% interest
B. 4 years at 2% interest
C. 4 years at 2.5% interest
D. 6 years at 1% interest
A. 3 years at 3% interest
B. 4 years at 2% interest
C. 4 years at 2.5% interest
D. 6 years at 1% interest
Answers
GPT-4o mini
To find the total cost of each loan option, we will use the formula \( A = P \left(1 + \frac{r}{n}\right)^{nt} \), where:
- \( A \) is the total amount after interest,
- \( P \) is the principal amount (the initial amount of the loan, $2,500),
- \( r \) is the annual interest rate (as a decimal),
- \( n \) is the number of times that interest is compounded per year (monthly, so \( n = 12 \)),
- \( t \) is the number of years.
Let's calculate the total amount for each option:
### Option A: 3 years at 3% interest
- \( P = 2500 \)
- \( r = 0.03 \)
- \( n = 12 \)
- \( t = 3 \)
\[
A = 2500 \left(1 + \frac{0.03}{12}\right)^{12 \times 3}
\]
\[
= 2500 \left(1 + 0.0025\right)^{36}
\]
\[
= 2500 \left(1.0025\right)^{36}
\]
\[
= 2500 \times 1.093443
\]
\[
A \approx 2733.61
\]
### Option B: 4 years at 2% interest
- \( P = 2500 \)
- \( r = 0.02 \)
- \( n = 12 \)
- \( t = 4 \)
\[
A = 2500 \left(1 + \frac{0.02}{12}\right)^{12 \times 4}
\]
\[
= 2500 \left(1 + 0.00166667\right)^{48}
\]
\[
= 2500 \left(1.00166667\right)^{48}
\]
\[
= 2500 \times 1.083282
\]
\[
A \approx 2708.20
\]
### Option C: 4 years at 2.5% interest
- \( P = 2500 \)
- \( r = 0.025 \)
- \( n = 12 \)
- \( t = 4 \)
\[
A = 2500 \left(1 + \frac{0.025}{12}\right)^{12 \times 4}
\]
\[
= 2500 \left(1 + 0.00208333\right)^{48}
\]
\[
= 2500 \left(1.00208333\right)^{48}
\]
\[
= 2500 \times 1.084046
\]
\[
A \approx 2710.12
\]
### Option D: 6 years at 1% interest
- \( P = 2500 \)
- \( r = 0.01 \)
- \( n = 12 \)
- \( t = 6 \)
\[
A = 2500 \left(1 + \frac{0.01}{12}\right)^{12 \times 6}
\]
\[
= 2500 \left(1 + 0.00083333\right)^{72}
\]
\[
= 2500 \left(1.00083333\right)^{72}
\]
\[
= 2500 \times 1.061677
\]
\[
A \approx 2654.19
\]
### Summary of Total Amounts:
- Option A: $2733.61
- Option B: $2708.20
- Option C: $2710.12
- Option D: $2654.19
### Conclusion
The loan term with the lowest total cost is **Option D: 6 years at 1% interest, with a total amount of approximately $2654.19.**
- \( A \) is the total amount after interest,
- \( P \) is the principal amount (the initial amount of the loan, $2,500),
- \( r \) is the annual interest rate (as a decimal),
- \( n \) is the number of times that interest is compounded per year (monthly, so \( n = 12 \)),
- \( t \) is the number of years.
Let's calculate the total amount for each option:
### Option A: 3 years at 3% interest
- \( P = 2500 \)
- \( r = 0.03 \)
- \( n = 12 \)
- \( t = 3 \)
\[
A = 2500 \left(1 + \frac{0.03}{12}\right)^{12 \times 3}
\]
\[
= 2500 \left(1 + 0.0025\right)^{36}
\]
\[
= 2500 \left(1.0025\right)^{36}
\]
\[
= 2500 \times 1.093443
\]
\[
A \approx 2733.61
\]
### Option B: 4 years at 2% interest
- \( P = 2500 \)
- \( r = 0.02 \)
- \( n = 12 \)
- \( t = 4 \)
\[
A = 2500 \left(1 + \frac{0.02}{12}\right)^{12 \times 4}
\]
\[
= 2500 \left(1 + 0.00166667\right)^{48}
\]
\[
= 2500 \left(1.00166667\right)^{48}
\]
\[
= 2500 \times 1.083282
\]
\[
A \approx 2708.20
\]
### Option C: 4 years at 2.5% interest
- \( P = 2500 \)
- \( r = 0.025 \)
- \( n = 12 \)
- \( t = 4 \)
\[
A = 2500 \left(1 + \frac{0.025}{12}\right)^{12 \times 4}
\]
\[
= 2500 \left(1 + 0.00208333\right)^{48}
\]
\[
= 2500 \left(1.00208333\right)^{48}
\]
\[
= 2500 \times 1.084046
\]
\[
A \approx 2710.12
\]
### Option D: 6 years at 1% interest
- \( P = 2500 \)
- \( r = 0.01 \)
- \( n = 12 \)
- \( t = 6 \)
\[
A = 2500 \left(1 + \frac{0.01}{12}\right)^{12 \times 6}
\]
\[
= 2500 \left(1 + 0.00083333\right)^{72}
\]
\[
= 2500 \left(1.00083333\right)^{72}
\]
\[
= 2500 \times 1.061677
\]
\[
A \approx 2654.19
\]
### Summary of Total Amounts:
- Option A: $2733.61
- Option B: $2708.20
- Option C: $2710.12
- Option D: $2654.19
### Conclusion
The loan term with the lowest total cost is **Option D: 6 years at 1% interest, with a total amount of approximately $2654.19.**