Math Problem Statement

Jonathan is buying a car today. The car costs R361731 and he can afford to make monthly payments of R12626 at the end of every month. He must pay interest at a rate of 7.065% per year compounded monthly. If the last payment Jonathan makes can be larger than R12626, how many payments must Jonathan make to repay the car loan?

Solution

To determine the number of payments Jonathan must make to repay the car loan, we can use the formula for the present value of an ordinary annuity. The formula is:

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

Where:

  • PVPV is the present value (the loan amount), which is R361731R361731.
  • PP is the monthly payment amount, which is R12626R12626.
  • rr is the monthly interest rate, calculated as the annual interest rate divided by 12.
  • nn is the number of payments.

Given:

  • The annual interest rate is 7.065%7.065\%, so the monthly interest rate rr is: r=7.065%12=0.07065120.0058875r = \frac{7.065\%}{12} = \frac{0.07065}{12} \approx 0.0058875

Let's solve for nn:

361731=12626×1(1+0.0058875)n0.0058875361731 = 12626 \times \frac{1 - (1 + 0.0058875)^{-n}}{0.0058875}

Now, let's calculate this equation step by step.The calculation shows that Jonathan needs to make approximately 31.47 payments. Since the number of payments must be an integer, this means he will make 31 regular payments of R12,626 each, and a final larger payment to cover the remaining balance.

Would you like more details on this calculation, or do you have any questions? Here are some related questions you might find useful:

  1. How much will the final payment be?
  2. How much interest will Jonathan pay in total?
  3. What would happen if the interest rate changed?
  4. How would the total number of payments change if Jonathan increased his monthly payment?
  5. What is the total cost of the car including interest?

Tip: It's useful to calculate the total interest paid over the life of a loan to understand the full cost of financing a purchase.