Question

Calculate the amount and compound interest on
(a) $Rs.10,800$ for $3$ years at $12\frac{1}{2}\%$ per annum compounded annually.
(b) $Rs.18,000$ for $2\frac{1}{2}$ years at $10\%$ per annum compounded annually.

Answer 1

We have, $A=P(1+\frac{r}{n\times 100}{)}^{nt}$ .......(i)
Where, $A$ is total amount.
$P$ is principal amount.
$t$ is time in years and
$n$ is number of times interest compounds annually.

(a) Given $P=Rs.10,800,t=3,R=12\frac{1}{2}=\frac{25}{2}\%$ and $n=1$ as interest compounds annually only one time.
Substituting given values in equation(i), we get

$A=10800{(1+\frac{25}{2\times 100})}^3$

$=10,800{(1+\frac{25}{200})}^3$
$=10,800(1+\frac{1}{8}{)}^3$

$=10,800{\left(\frac{9}{8}\right)}^3$

$=10,800\times \frac{9}{8}\times \frac{9}{8}\times \frac{9}{8}$

$=15377.34$

Therefore, the total amount after $3$ years is $Rs.15377.34$

So, compound interest $I=A-P=Rs.15377.34-Rs.10,800=Rs.4577.34$

Hence, compound interest is $Rs.4577.34$

(b) Given $p=Rs.18,000,t=\frac{5}{2},r=10$ and $n=1$

Substituting given values in equation(i) for $2$ years,
we get

$A=P{(1+\frac{R}{100})}^2$

$=18,000{(1+\frac{10}{100})}^2$

$=18,000{\left(\frac{11}{10}\right)}^2$

$=18,000\times \frac{11}{10}\times \frac{11}{10}$

$=21780$

So, the Interest after $2$ year is $Rs.21780-Rs.18000$

$=Rs.3780$

And interest after half year is:

$=\frac{PNR}{100}$

$=21780\times \frac{10}{200}$

$=Rs.1080$

Therefore, total interest $=Rs.3780+Rs.1080=Rs.4860$

and the total amount after $2\frac{1}{2}$ years is $Rs.21780+Rs.1080=Rs.22,860$