Math Problem Statement

The cash price of a car is 970,000 Php. The dealer offers you a loan arrangement of 22,339.50 Php per month payable for 5 years. How much interest is being imposed on this loan? Make a table of f(x) for intervals of 2% for x to determine the initial range of the root. Then, using this initial range, apply false-position method for 3 iterations to approach the root.

Solution

The problem involves determining the interest rate on a car loan given monthly payments. We'll treat this as a problem of solving for the interest rate xx in the context of a loan. The monthly payment formula for an amortized loan is:

P=A(x12)1(1+x12)nP = \frac{A \cdot \left(\frac{x}{12}\right)}{1 - \left(1 + \frac{x}{12}\right)^{-n}}

Where:

  • P=22,339.50P = 22,339.50 Php (monthly payment)
  • A=970,000A = 970,000 Php (loan amount or principal)
  • n=5×12=60n = 5 \times 12 = 60 (total number of payments, as 5 years = 60 months)
  • xx is the annual interest rate (what we are solving for).

We will:

  1. Create a function f(x)f(x) that represents the difference between the monthly payment formula and the actual payment.
  2. Create a table of f(x)f(x) for different interest rates in 2% intervals to determine the initial range of the root.
  3. Use the false-position method to refine the solution for xx, performing 3 iterations.

The function f(x)f(x) is:

f(x)=A(x12)1(1+x12)6022,339.50f(x) = \frac{A \cdot \left(\frac{x}{12}\right)}{1 - \left(1 + \frac{x}{12}\right)^{-60}} - 22,339.50

I'll compute the values for different xx intervals and apply the false-position method for 3 iterations. Let's begin by calculating the values of f(x)f(x) for the interest rate intervals.

Here is the table of f(x)f(x) values for interest rates at 2% intervals:

\hline \text{Interest Rate (x)} & f(x) \\ \hline 0.02 & -5337.57 \\ 0.04 & -4475.47 \\ 0.06 & -3586.68 \\ 0.08 & -2671.40 \\ 0.10 & -1729.87 \\ 0.12 & -762.39 \\ 0.14 & 230.70 \\ 0.16 & 1249.02 \\ 0.18 & 2292.12 \\ \hline \end{array}$$ From this table, the root lies between 12% and 14%, as the function changes sign in this interval. Now, we will apply the **false-position method** using this initial range. ### Steps of False-Position Method: 1. Identify $$x_0 = 0.12$$ and $$x_1 = 0.14$$ as the initial range where the root lies. 2. Use the formula to compute the next approximation: $$x_{\text{new}} = x_1 - \frac{f(x_1) \cdot (x_1 - x_0)}{f(x_1) - f(x_0)}$$ Let's proceed with 3 iterations of the false-position method.The results of applying the false-position method for 3 iterations are as follows: $$\begin{array}{|c|c|c|c|c|} \hline \text{Iteration} & x_0 & x_1 & x_{\text{new}} & f(x_{\text{new}}) \\ \hline 1 & 0.12 & 0.14 & 0.13535 & -2.264 \\ 2 & 0.13535 & 0.14 & 0.13540 & -0.00657 \\ 3 & 0.13540 & 0.14 & 0.13540 & -0.00002 \\ \hline \end{array}$$ After 3 iterations, the estimated interest rate is approximately **13.54%**. ### Conclusion: The interest being imposed on this loan is about **13.54%**. Would you like more details, or do you have any questions? Here are some related questions: 1. How does the false-position method compare to other root-finding methods? 2. What factors affect the accuracy of the false-position method? 3. How does compound interest influence loan payments? 4. Can interest rates vary during the loan period? 5. How can overpayment impact the total interest paid on a loan? **Tip:** For faster convergence in the false-position method, choose initial points that are close to the root.

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Math Problem Analysis

Mathematical Concepts

Amortized Loan
Root-Finding
False-Position Method

Formulas

Monthly Payment Formula: P = A * (x/12) / (1 - (1 + x/12)^-n)
False-Position Method: x_new = x1 - (f(x1) * (x1 - x0)) / (f(x1) - f(x0))

Theorems

Amortization Formula
False-Position Method

Suitable Grade Level

College Level, Advanced High School (Grades 11-12)