Question 1193612
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 (\( 2 \, \text{years} \times 12 = 24 \))  

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### 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:

\[
3,200 \cdot \left( (1 + r)^{24} - 1 \right) = 60,000 \cdot r \cdot (1 + r)^{24}
\]

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### 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:

\[
0.05333 = \frac{r \cdot (1 + r)^{24}}{(1 + r)^{24} - 1}
\]

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### 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:

\[
r = 0.01 \, \text{(or 1% per month)}.
\]

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### Step 4: Convert to Annual Interest Rate

The nominal annual interest rate is:

\[
\text{Annual Rate} = r \cdot 12 = 0.01 \cdot 12 = 0.12 \, \text{(or 12% per year)}.
\]

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### Final Answer:
The interest rate charged on the loan is **12% per year compounded monthly**.