Question

Calculate the amount and compound interest on Rs.15000 for 2 years at 10% per annum compounded annually.

Answer 1

Hint: First we find out the amount generated $A$ after 2 years using the compound interest formula $A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}$ where the principal is $P=15000$ rupees, rate of interest $r=10%$, compound frequency $n=1$ and the time period $t=2$ years. Then we find a compound which is the difference between the amount generated and the principal.

Complete step-by-step answer:
The compound interest is the concept where the interest is added into the principal and in the next compound period the new principal is the sum of old principal and accumulated interest. \[\]
We shall denote the original principal sum as $P$, the overall compound period as $t$, the rate of interest per time period as $r$, the number of compound periods or frequency as $n$ and the amount accumulated after the time period $nt$ as $A$. Then we know from the formula of compound interest,
\[A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\]
The compound interest $I$ generated after time period $nt$ is given by
\[I=A-P\]

We find from the given data in the question that we have to calculate compound interest $I$. We first need to find the accumulated amount $A$. We put the principal $P=150000$, the rate of interest $r=10%=0.01$, the number of compound periods $n=1$ year and finally the overall time period $t=2$ in the compound interest formula to get,
\[A=15000{{\left( 1+\dfrac{0.1}{1} \right)}^{1\times 2}}=15000\times {{1.1}^{2}}=18510\text{Rs}\]
The compound interest $I$ generated after time period 2 years is ,
\[I=18510-15000=3510\text{Rs}\]

Note: We need to be careful of the difference between compound interest and simple interest whose formula is $A=P\left( 1+rt \right)$ where the compound frequency is always 1. We use the formula $P\left( t \right)={{P}_{0}}{{e}^{rt}}$ (${{P}_{0}}$ is the initial amount and $e$ is the base of natural logarithm) when the compound period is infinitely small .