Question
Jocelyn is considering taking out one of the two following loans. Loan H is a three-year loan with a principal of $5,650 and an interest rate of 12.24%, compounded monthly. Loan I is a four-year loan with a principal of $6,830 and an interest rate of 10.97%, compounded monthly. Which loan will have the smaller monthly payment, and how much smaller will it be? Round all dollar values to the nearest cent.
a.
Loan H's monthly payment will be $42.46 smaller than Loan I's.
b.
Loan H's monthly payment will be $140.79 smaller than Loan I's.
c.
Loan I's monthly payment will be $11.88 smaller than Loan H's.
d.
Loan I's monthly payment will be $26.98 smaller than Loan H's.
Please select the best answer from the choices provided
A
B
C
D
a.
Loan H's monthly payment will be $42.46 smaller than Loan I's.
b.
Loan H's monthly payment will be $140.79 smaller than Loan I's.
c.
Loan I's monthly payment will be $11.88 smaller than Loan H's.
d.
Loan I's monthly payment will be $26.98 smaller than Loan H's.
Please select the best answer from the choices provided
A
B
C
D
Answers
Answered by
GPT-4o mini
To compare the monthly payments of Loan H and Loan I, we will use the formula for the monthly payment on an amortizing loan, which is:
\[
M = P \frac{r(1 + r)^n}{(1 + r)^n - 1}
\]
Where:
- \( M \) = monthly payment
- \( P \) = principal amount (loan amount)
- \( r \) = monthly interest rate (annual rate / 12)
- \( n \) = total number of payments (loan term in months)
**Calculating Loan H:**
- Principal (P) = $5,650
- Annual Interest Rate = 12.24%
- Monthly Interest Rate (r) = 12.24% / 100 / 12 = 0.0102
- Loan Term = 3 years = 36 months (n)
Now, plug the values into the formula:
\[
M_H = 5650 \frac{0.0102 (1 + 0.0102)^{36}}{(1 + 0.0102)^{36} - 1}
\]
First, calculate \((1 + 0.0102)^{36}\):
\[
(1 + 0.0102)^{36} \approx (1.0102)^{36} \approx 1.432364654
\]
Now, substitute this value back into the formula:
\[
M_H = 5650 \frac{0.0102 \times 1.432364654}{1.432364654 - 1}
\]
\[
M_H \approx 5650 \frac{0.014606122}{0.432364654}
\]
\[
M_H \approx 5650 \times 0.0338 \approx 191.83
\]
**Calculating Loan I:**
- Principal (P) = $6,830
- Annual Interest Rate = 10.97%
- Monthly Interest Rate (r) = 10.97% / 100 / 12 = 0.00914
- Loan Term = 4 years = 48 months (n)
Calculate \((1 + 0.00914)^{48}\):
\[
(1 + 0.00914)^{48} \approx (1.00914)^{48} \approx 1.456205628
\]
Now, substitute this value back into the formula:
\[
M_I = 6830 \frac{0.00914 \times 1.456205628}{1.456205628 - 1}
\]
\[
M_I \approx 6830 \frac{0.013321345}{0.456205628}
\]
\[
M_I \approx 6830 \times 0.0292 \approx 199.65
\]
**Finding the difference in monthly payments:**
Now to find the difference between the two monthly payments:
\[
\text{Difference} = M_I - M_H \approx 199.65 - 191.83 \approx 7.82
\]
Thus, the monthly payment for Loan H is less than Loan I by about $7.82.
Among the provided options, we can conclude that none of the options exactly match our calculations. However, the correct interpretation from this data is that Loan H does have a smaller payment, but the exact difference is about $7.82, which is significantly lower than the options provided.
Thus, none of the presented answers accurately represent the calculated difference.
\[
M = P \frac{r(1 + r)^n}{(1 + r)^n - 1}
\]
Where:
- \( M \) = monthly payment
- \( P \) = principal amount (loan amount)
- \( r \) = monthly interest rate (annual rate / 12)
- \( n \) = total number of payments (loan term in months)
**Calculating Loan H:**
- Principal (P) = $5,650
- Annual Interest Rate = 12.24%
- Monthly Interest Rate (r) = 12.24% / 100 / 12 = 0.0102
- Loan Term = 3 years = 36 months (n)
Now, plug the values into the formula:
\[
M_H = 5650 \frac{0.0102 (1 + 0.0102)^{36}}{(1 + 0.0102)^{36} - 1}
\]
First, calculate \((1 + 0.0102)^{36}\):
\[
(1 + 0.0102)^{36} \approx (1.0102)^{36} \approx 1.432364654
\]
Now, substitute this value back into the formula:
\[
M_H = 5650 \frac{0.0102 \times 1.432364654}{1.432364654 - 1}
\]
\[
M_H \approx 5650 \frac{0.014606122}{0.432364654}
\]
\[
M_H \approx 5650 \times 0.0338 \approx 191.83
\]
**Calculating Loan I:**
- Principal (P) = $6,830
- Annual Interest Rate = 10.97%
- Monthly Interest Rate (r) = 10.97% / 100 / 12 = 0.00914
- Loan Term = 4 years = 48 months (n)
Calculate \((1 + 0.00914)^{48}\):
\[
(1 + 0.00914)^{48} \approx (1.00914)^{48} \approx 1.456205628
\]
Now, substitute this value back into the formula:
\[
M_I = 6830 \frac{0.00914 \times 1.456205628}{1.456205628 - 1}
\]
\[
M_I \approx 6830 \frac{0.013321345}{0.456205628}
\]
\[
M_I \approx 6830 \times 0.0292 \approx 199.65
\]
**Finding the difference in monthly payments:**
Now to find the difference between the two monthly payments:
\[
\text{Difference} = M_I - M_H \approx 199.65 - 191.83 \approx 7.82
\]
Thus, the monthly payment for Loan H is less than Loan I by about $7.82.
Among the provided options, we can conclude that none of the options exactly match our calculations. However, the correct interpretation from this data is that Loan H does have a smaller payment, but the exact difference is about $7.82, which is significantly lower than the options provided.
Thus, none of the presented answers accurately represent the calculated difference.