Math Problem Statement

Dr. & Mr. Wilson were delighted to find a house for sale that they loved which required them to borrow $160,000. They need to decide between a 20-year or 30-year loan. Both loans charge interest at a rate of 9.6% per year compounded monthly. 20-year Loan (a.) Find the monthly payments for the 20-year loan.

Solution

To calculate the monthly payment for the 20-year loan, we can use the formula for the monthly payment on an amortized loan:

M=P×r(1+r)n(1+r)n1M = P \times \frac{r(1 + r)^n}{(1 + r)^n - 1}

Where:

  • MM = monthly payment
  • PP = loan principal (the amount borrowed) = $160,000
  • rr = monthly interest rate (annual rate divided by 12)
  • nn = total number of payments (loan term in years multiplied by 12)

Given:

  • Annual interest rate = 9.6% = 0.096
  • Loan term = 20 years
  • Monthly interest rate r=9.6%12=0.09612=0.008r = \frac{9.6\%}{12} = \frac{0.096}{12} = 0.008
  • Total number of payments n=20×12=240n = 20 \times 12 = 240

Let's compute the monthly payment using these values.The monthly payment for the 20-year loan would be approximately $1,501.87.

Do you have any questions or would you like further details on this?

Here are 5 related questions to consider:

  1. How much interest will be paid over the life of the 20-year loan?
  2. How do monthly payments differ if the loan term is increased to 30 years?
  3. How does a change in interest rate affect monthly payments?
  4. What is the total cost of the house with interest included for the 20-year loan?
  5. What is the benefit of paying off a loan in 20 years versus 30 years?

Tip: Making extra payments on a loan can significantly reduce the total interest paid over the life of the loan.

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Math Problem Analysis

Mathematical Concepts

Amortized Loan Payments
Compound Interest
Financial Mathematics

Formulas

M = P * (r(1 + r)^n) / ((1 + r)^n - 1)
r = annual interest rate / 12
n = loan term in years * 12

Theorems

Loan Amortization Formula

Suitable Grade Level

College/University Level (Finance or Business Mathematics)