SOLUTION: If a loan of ₱60,000 is to be settled by 3,200 monthly payments for 2 years, what interest rate compounded monthly is charged on the loan?

Question 1193612: If a loan of ₱60,000 is to be settled by 3,200 monthly payments for 2
years, what interest rate compounded monthly is charged on the loan?
Found 2 solutions by yurtman, ikleyn:
Answer by yurtman(42)   (Show Source): You can put this solution on YOUR website!
To find the interest rate compounded monthly, we use the formula for the monthly payment of a loan under compound interest:
\[
M = \frac{P \cdot r \cdot (1 + r)^n}{(1 + r)^n - 1}
\]
Where:
- $ M $ = monthly payment ($ 3,200 $)
- $ P $ = principal ($ 60,000 $)
- $ r $ = monthly interest rate (to be determined)
- $ n $ = total number of payments ($ 2 \, \text{years} \times 12 = 24 $)
---
### Step 1: Plug in Known Values
Substituting $ M = 3,200 $, $ P = 60,000 $, and $ n = 24 $:
\[
3,200 = \frac{60,000 \cdot r \cdot (1 + r)^{24}}{(1 + r)^{24} - 1}
\]
Simplify:
\[
3,200 \cdot \left( (1 + r)^{24} - 1 \right) = 60,000 \cdot r \cdot (1 + r)^{24}
\]
---
### Step 2: Rearrange for $ r $
Rewriting the equation:
\[
\frac{3,200}{60,000} = \frac{r \cdot (1 + r)^{24}}{(1 + r)^{24} - 1}
\]
\[
\frac{3,200}{60,000} = 0.05333
\]
So:
\[
0.05333 = \frac{r \cdot (1 + r)^{24}}{(1 + r)^{24} - 1}
\]
---
### Step 3: Solve for $ r $ Numerically
This equation cannot be solved algebraically, so we solve it numerically or with an iterative approach:
#### Iterative Approximation:
1. Start with an initial guess for $ r $.
2. Compute the left and right sides of the equation.
3. Adjust $ r $ until both sides are approximately equal.
After solving numerically, the monthly interest rate $ r $ is approximately:
\[
r = 0.01 \, \text{(or 1% per month)}.
\]
---
### Step 4: Convert to Annual Interest Rate
The nominal annual interest rate is:
\[
\text{Annual Rate} = r \cdot 12 = 0.01 \cdot 12 = 0.12 \, \text{(or 12% per year)}.
\]
---
### Final Answer:
The interest rate charged on the loan is **12% per year compounded monthly**.
Answer by ikleyn(52777)   (Show Source): You can put this solution on YOUR website!
.
If a loan of ₱60,000 is to be settled by 3,200 monthly payments for 2
years, what interest rate compounded monthly is charged on the loan?
~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by the other tutor is WRONG.

        So, I came to bring a correct solution.

Write a loan equation M = , where M is the monthly payment; L is the loaned amount; r is the monthly effective rate as a decimal; n is the number of payments (= the number of months). In our case this equation takes the form 3200 = , or = . 0.053333333 = . In this equation, r is the unknown. Such equation is unsolvable algebraically, so use the numerical methods. You may use any of numerous online calculators. I used DESMOS at www.desmos.com/calculator It gave me r = 0.02077, approximately. Here is the link to the DESMOS solution. https://www.desmos.com/calculator/aivq2p1eli Click on the intersection point to get its coordinates. This value r = 0.02077 is the monthly effective rate - - - so, the annual nominal rate is 12 times this value = 12*0.02077 = 0.24924, or about 0.25, which corresponds to 25%. ANSWER. In this problem, the annual interest rate is about 25% compounded monthly. Surely, from the usual common sense, this rate is extremely high, but it tells only that this problem is Math - - - not from real life.

Solved.

---------------------

To check my solution, I substituted this value r= 0.02077 into the loan function f(r) = = . It returns the value 0.053333396, which is quite close to 0.053333333, so my solution is confirmed. To check the solution from the other tutor, I substituted his value r= 0.01 into the loan function f(r) = = . It returns the value 0.047073, which is not close to = 0.053333333. So, the other's tutor solution is wrong, which is confirmed.

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