Question

Find the compound interest on Rs 4000 for $2\frac{1}{2}$ years at 10%per annum compounded annually

Answer 1

$P=4000$

$R=10\%$

Times $=2.5$ years

First take $n=2$ years

$A=P{(1+\frac{r}{100})}^n=4000{\left(\frac{11}{10}\right)}^2$

$4000\left(\frac{121}{100}\right)$

$A=4840$

Simple interest for last $\frac{1}{2}$ years is $\frac{PRT}{100}$

$=\frac{4840\times 10}{100}\times \frac{1}{2}$

$=242$

Amount after $2\frac{1}{2}$ year $=4840+242$

$=5082$

Compound interest $=5082-4000$

CI$=1082$.

Hence, the compound interest, CI$=1082$