Question

A sum of $9600$ is invested for $3$ years at $10\%$ per annum at compound interest.
(i) What is the sum due at the end of the first year?
(ii) What is the sum due at the end of the second year?
(iii) Find the compound interest earned in $2$ years.
(iv) Find the difference between the answers in (ii) and (i) and find the interest on this

Answer 1

Since, Interest for the $1^{st}$ year $\frac{P\times T\times R}{100}$
$=Rs\frac{9600\times 10\times 1}{100}=Rs960$

Amount after the $1^{st}$ year $=Rs9600+Rs960=Rs10560$

$\left(i\right)$ The sum due at the end of the $1^{st}$ year $=Rs10560$

Interest for the $2^{nd}$ year $=Rs\frac{10560\times 10\times 1}{100}=Rs1056$

Amount after the $2^{nd}$ year $=Rs10560+Rs1056=Rs11616$

$\left(ii\right)$ The sum due at the end of the $2^{nd}$ year $=Rs11616$

$\left(iii\right)$ Compound Interest earned in $2$ years $=Rs11616-Rs9600=Rs2016$

$\left(iv\right)$ Difference $=Rs11616-Rs10560=Rs.1056$

Interest $=\frac{1056\times 3\times 10}{100}=Rs316.80$