Question

At an annual interest rate of 2.5%, an original principal can accrue an interest of K350.25 over an 18 month's period. Calculate the principal if interest is to be compounded every three months.

Answer 1

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:

\[
\boxed{9189.23}
\]