Interesting interests

Let $\text{SI}$ be the simple interest, $\text{R}$ be the rate of interest, $\text{n}$ be the time period, $\text{P}$ be the principal, ${\text{A}}_1$ be the simple interest amount and ${\text{A}}_2$ be the compound interest amount, then:-

When the interest is calculated by simple interest method:-
$\begin{aligned} \text{SI} & = \dfrac{\text{PRn}}{100}\\ \\ & = \dfrac{360 × 9 × 2}{100}\\ \\ & = 64.8\\ \\ \implies {\text{A}}_1 & = \text{P} + \text{SI}\\ & = 360 + 64.8\\ & = 424.8 \end{aligned}$

When interest is compounded annually:-
$\begin{aligned} {\text{A}}_2 & = \text{P}{\left(1 + \dfrac{\text{R}}{100}\right)}^{\text{n}}\\ \\ & = 360 × {\left(1 + \dfrac{9}{100}\right)}^{2}\\ \\ & = 360 × {\left(\dfrac{109}{100}\right)}^{2}\\ \\ & = 360 × \dfrac{109}{100} × \dfrac{109}{100}\\ \\ & = 427.716 \end{aligned}$

So, Abhay would have earned $\text{A}_2 - \text{A}_1$ more if the interest was compounded annually.
So, the answer is $1000\left(427.716 - 424.8\right) = 1000 × 2.916 = \color{#3D99F6}{\boxed{2916}}$ .

• On simple interest, principal and interest after two years $= Rs.360 \times (1+0.09\times 2) = Rs.424.8$
• On compound interest, principal and interest after two years $= Rs.360 \times (1+0.09)^2 = Rs.427.716$
• Extra payment of interest $A = 427.716 - 424.8 = 2.916$
• $\implies 1000A = \boxed{2916}$