Math Problem Statement
A sailboat costs
$ 19 comma 931$19,931.
You pay
10%
down and amortize the rest with equal monthly payments over a
15-year
period. If you must pay
8.1 %
compounded monthly, what is your monthly payment? How much interest will you pay?
Solution
To calculate the monthly payment and total interest for this loan, we will use the amortization formula:
Where:
- is the monthly payment,
- is the loan principal (amount borrowed),
- is the monthly interest rate,
- is the total number of payments (number of months).
Step 1: Calculate the Principal
You are making a 10% down payment, so the remaining balance is 90% of the cost:
Step 2: Convert the Interest Rate
The annual interest rate is 8.1%, but since it's compounded monthly, we convert it to a monthly rate:
Step 3: Calculate the Total Number of Payments
The loan is amortized over 15 years, so the total number of monthly payments is:
Step 4: Calculate the Monthly Payment and Total Interest
Now, let's compute the monthly payment and the total interest paid over the life of the loan.The monthly payment for the sailboat is approximately $172.46.
The total interest paid over the 15-year period will be approximately $13,105.09.
Would you like further clarification or another example?
Here are some related questions to explore:
- How much would the monthly payment be with a 5% interest rate?
- What happens if the loan period is shortened to 10 years?
- How much total interest would you save if you paid an extra $50 per month?
- How does changing the down payment affect the monthly payment?
- What if the interest were compounded quarterly instead of monthly?
Tip: Reducing the loan term or making larger down payments can significantly reduce the total interest paid.
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Math Problem Analysis
Mathematical Concepts
Loan Amortization
Interest Rate Conversion
Compound Interest
Formulas
Amortization Formula: M = P * [r(1 + r)^n] / [(1 + r)^n - 1]
Interest Rate Conversion: r = Annual Rate / 12
Theorems
Amortization of Loans
Suitable Grade Level
College Level or Advanced High School (Grades 11-12)
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