Question

A man invests Rs. $\mathrm{9,600}$ at $10\%$ per annum compound interest for $3$ years. Find the interest for the third year.

Answer 1

Simple Interest $I=\frac{PNR}{100}$
Here for first year,
$P=$ Rs. $\mathrm{9,600}$, $N=1$ year, $R=10\%$
So, Interest at the end of 1st year $I=\frac{9,600\times 1\times 10}{100}=$ Rs. $960$

And for second year,
$P=$ Rs. $9,600+$ Rs. $960=$ Rs. $10,560$
$N=1$ year, $R=10\%$
$I=\frac{10,560\times 1\times 10}{100}=$ Rs. $1,056$
And for third year,
$P=$ Rs. $10,560+$ Rs. $1,056=$ Rs. $11,616$
$N=1$ year, $R=10\%$
$I=\frac{11,616\times 1\times 10}{100}=$ Rs. $1,161.60$

Therefore, the interest for the third years is Rs. $1,161.60$.