Amortization is a key financial concept that helps students connect math to real-world scenarios like buying cars, paying for college, or getting a home loan. Use these word problems to teach students about loans, interest, and repayment calculations. Each question includes a step-by-step solution, making them perfect for classroom use, homework, or group projects.
Problem 1: Basic Fixed-Rate Loan
Question:
Samantha takes out a $5,000 loan to buy a used car. The loan has a fixed annual interest rate of 6% and must be paid back in monthly installments over 3 years. What will Samantha’s monthly payment be? (Round to the nearest cent.)
Solution:
Principal (P) = $5,000
Annual rate (r) = 6% = 0.06
Monthly rate = 0.06 / 12 = 0.005
Number of months (n) = 3 × 12 = 36
Monthly payment formula:
A = [P × r × (1 + r)n] / [(1 + r)n – 1]
A = [5000 × 0.005 × (1.005)36] / [(1.005)36 – 1]
A ≈ $152.10
Answer: Samantha’s monthly payment will be $152.10.
Problem 2: Extra Payments
Question:
Jorge has a $10,000 student loan at 5% annual interest, to be repaid over 5 years (60 months). If he pays an extra $50 each month on top of his regular payment, how much interest will he save compared to making just the required payments?
Solution Outline:
Find the regular monthly payment using the amortization formula. Calculate total interest paid with and without extra payments (can be done as a spreadsheet activity or with an online calculator). Discuss how extra payments reduce total interest.
Key Concept: Extra payments reduce the principal faster, so less interest accumulates over time.
Problem 3: Comparing Loan Terms
Question:
Maria is considering two loans for $20,000:
– Loan A: 5-year term at 4% interest
– Loan B: 10-year term at 4% interest
What are the monthly payments for each, and how much total interest will she pay under each plan?
Solution:
For each loan, use the amortization formula.
Loan A (5 years):
n = 60
r = 0.04 / 12 = 0.00333
A ≈ $368.33/month
Total paid: $368.33 × 60 = $22,099.80
Total interest: $2,099.80
Loan B (10 years):
n = 120
A ≈ $202.49/month
Total paid: $202.49 × 120 = $24,298.80
Total interest: $4,298.80
Answer:
– Loan A: $368.33/month, $2,099.80 total interest
– Loan B: $202.49/month, $4,298.80 total interest
Problem 4: Understanding the Impact of Rates
Question:
A $2,000 loan can be paid off in 2 years at 5% annual interest, or 2 years at 10% annual interest. What are the monthly payments for each option? Which loan costs more in interest?
Solution: Calculate the payment for each scenario using the amortization formula and compare total interest paid.
Problem 5: Amortization Table Activity
Activity: Provide students with the first five rows of an amortization schedule (see example below).
Ask students to:
- Identify which columns represent principal and interest
- Calculate the remaining balance after each payment
- Observe the trend in how principal and interest portions change over time
| Payment # | Payment Amt | Interest Paid | Principal Paid | Balance |
|---|---|---|---|---|
| 1 | $100 | $10 | $90 | $910 |
| 2 | $100 | $9.10 | $90.90 | $819.10 |
| 3 | $100 | $8.19 | $91.81 | $727.29 |
| 4 | $100 | $7.27 | $92.73 | $634.56 |
| 5 | $100 | $6.35 | $93.65 | $540.91 |