nd the amount and the compound interest on \( \mathrm{N} 1,00,000 \) compounde uarterly for 9 months at the rate of \( 4 \% \) per annum.

To find the amount and the compound interest on \( \mathrm{N} 1,00,000 \) compounded quarterly for 9 months at the rate of \( 4 \% \) per annum, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount after the given time period - \( P \) is the principal amount (initial amount) - \( r \) is the annual interest rate (in decimal form) - \( n \) is the number of times the interest is compounded per year - \( t \) is the time period in years Given: - Principal amount (\( P \)) = \( \mathrm{N} 1,00,000 \) - Annual interest rate (\( r \)) = \( 4 \% \) = \( 0.04 \) - Compounded quarterly, so \( n = 4 \) - Time period (\( t \)) = 9 months = \( \frac{9}{12} \) years = \( \frac{3}{4} \) years Substitute the given values into the formula to find the amount and the compound interest. Calculate the value by following steps: - step0: Calculate: \(100000\left(1+\frac{0.04}{4}\right)^{\frac{4\times 3}{4}}\) - step1: Reduce the fraction: \(100000\left(1+\frac{0.04}{4}\right)^{3}\) - step2: Divide the terms: \(100000\left(1+\frac{1}{100}\right)^{3}\) - step3: Add the numbers: \(100000\left(\frac{101}{100}\right)^{3}\) - step4: Simplify: \(100000\times \frac{101^{3}}{100^{3}}\) - step5: Rewrite the expression: \(10^{5}\times \frac{101^{3}}{10^{6}}\) - step6: Reduce the numbers: \(1\times \frac{101^{3}}{10}\) - step7: Multiply: \(\frac{101^{3}}{10}\) The amount after 9 months at a compound interest rate of \( 4 \% \) per annum compounded quarterly is \( \frac{101^{3}}{10} \) or approximately \( 103030.1 \). To find the compound interest, we subtract the principal amount from the amount: Compound Interest = Amount - Principal Amount = \( 103030.1 - 100000 \) = \( 3030.1 \) Therefore, the compound interest on \( \mathrm{N} 1,00,000 \) compounded quarterly for 9 months at the rate of \( 4 \% \) per annum is \( 3030.1 \).