To find the principal when interest is compounded quarterly, we can use the formula for compound interest:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
where:
- \( A \) is the total amount after time \( t \).
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual nominal interest rate (as a decimal).
- \( n \) is the number of times that interest is compounded per unit \( t \).
- \( t \) is the time the money is invested or borrowed for, in years.
Given:
- The annual interest rate \( r = 2.5% = 0.025 \)
- The time period \( t = 18 \text{ months} = 1.5 \text{ years} \)
- Interest accrued \( A - P = K350.25 \)
Since we know the interest is compounded quarterly, we have:
- \( n = 4 \) (because interest is compounded quarterly)
From the information given, we can write:
\[ A = P + 350.25 \]
Replacing \( A \) in the compound interest formula:
\[ P + 350.25 = P \left(1 + \frac{0.025}{4}\right)^{4 \times 1.5} \]
First, let’s calculate the interest rate per quarter:
\[ \frac{0.025}{4} = 0.00625 \]
Now, we can calculate \( n \cdot t \):
\[ n \cdot t = 4 \times 1.5 = 6 \]
Now substitute these values into the equation:
\[ P + 350.25 = P \left(1 + 0.00625\right)^{6} \]
Calculating the expression inside the parentheses:
\[ 1 + 0.00625 = 1.00625 \]
Now, calculate \( (1.00625)^{6} \):
\[ (1.00625)^{6} \approx 1.0381 \quad \text{(using a calculator)} \]
Now substitute this back into the equation:
\[ P + 350.25 = P \times 1.0381 \]
Rearranging gives:
\[ P \times 1.0381 - P = 350.25 \]
Factoring out \( P \):
\[ P(1.0381 - 1) = 350.25 \]
This simplifies to:
\[ P(0.0381) = 350.25 \]
Now, solve for \( P \):
\[ P = \frac{350.25}{0.0381} \approx 9,189.23 \]
Thus, the principal \( P \) is approximately:
\[ \boxed{9189.23} \]