Question

For an investment, if the nominal rate of interest is\(10\%\)compounded half-yearly, then the effective rate of interest is:

• A. \(10.25\%\)
• B. \(11.25\%\)
• C. \(10.125\%\)
• D. \(11.025\%\)

Hint:

To calculate the effective interest rate, remember to account for the compounding frequency. The formula \( \text{Effective Rate} = \left( 1 + \frac{r}{n} \right)^n - 1 \) takes into account how often interest is compounded within a year. The more frequently the interest is compounded (i.e., larger values of \( n \)), the higher the effective rate will be, even if the nominal rate remains constant. Always ensure you substitute the values for \( r \) and \( n \) correctly and simplify step by step.

Answer 1

The Correct Option is A

The formula for the effective rate of interest is:
\(\text{Effective Rate} = \left( 1 + \frac{r}{n} \right)^n - 1,\)
where \(r = 0.1\) (nominal rate) and \(n = 2\) (compounding frequency per year).

Substitute into the formula:
\(\text{Effective Rate} = \left( 1 + \frac{0.1}{2} \right)^2 - 1 = \left( 1 + 0.05 \right)^2 - 1.\)

Simplify:
\(\text{Effective Rate} = (1.05)^2 - 1 = 1.1025 - 1 = 0.1025 = 10.25\%.\)

Thus, the effective rate of interest is \(10.25\%\).

Answer 2

The formula for the effective rate of interest is:

\[ \text{Effective Rate} = \left( 1 + \frac{r}{n} \right)^n - 1, \] where \( r = 0.1 \) (the nominal interest rate) and \( n = 2 \) (the compounding frequency per year, meaning the interest is compounded semi-annually).

Step 1: Substitute the given values into the formula:

Substituting \( r = 0.1 \) and \( n = 2 \) into the formula: \[ \text{Effective Rate} = \left( 1 + \frac{0.1}{2} \right)^2 - 1 = \left( 1 + 0.05 \right)^2 - 1. \]

Step 2: Simplify the expression:

Now, simplify: \[ \text{Effective Rate} = (1.05)^2 - 1 = 1.1025 - 1 = 0.1025. \]

Step 3: Convert to percentage:

Finally, converting the decimal into a percentage: \[ 0.1025 = 10.25\%. \]

Conclusion: The effective rate of interest is \( 10.25\% \).