Question

The compound interest on $Rs.4000$ at $10\%$ per annum for $2$ years $3$ months, Find compounded annually.

Answer 1

Principal $\left(P\right)=Rs.4000$

Rate of Interest $\left(r\right)=10\%$

Time $=2$ years and $3$ months

For first $2$ years:

$n=2$ years

Amount on compound interest,

$A=P{(1+\frac{r}{100})}^n$

$\Rightarrow A=4000{\left(\frac{11}{10}\right)}^2$

$\Rightarrow A=4000\times \frac{121}{100}$

$\Rightarrow A=Rs.4840$

The Amount after two years $=Rs.4840$

Now, for last $3$ months,i.e. for $\frac{1}{4}$ years:

Principal $=Rs.4840$

Simple interest

$=\frac{PRT}{100}=\frac{4840\times 10\times \frac{1}{4}}{100}=Rs.121$

Therefore,

Amount after $2$ years and $3$ months $=Rs.(4840+121)=Rs.4961$

Therefore,

$C.I.=A-P$

$\Rightarrow C.I.=Rs.(4961-4000)=Rs.961$

Hence the compound interest is $Rs.961$.