A sum of ₹10,000 amounts to ₹12,321 in 2 years at a certain rate of interest per annum, compounded annually. The rate of interest per annum is:
This problem asks us to find the annual rate of interest when a principal amount grows to a certain value over a period, with interest compounded annually.
Compound interest means that the interest earned in each period is added to the principal amount for the next period. The formula for the amount (A) when interest is compounded annually is:
$A = P \left(1 + \frac{R}{100}\right)^n$
Where:
From the question, we have:
$12,321 = 10,000 \left(1 + \frac{R}{100}\right)^2$
$\frac{12,321}{10,000} = \left(1 + \frac{R}{100}\right)^2$
$1.2321 = \left(1 + \frac{R}{100}\right)^2$
$\sqrt{1.2321} = 1 + \frac{R}{100}$
We know that $111 \times 111 = 12321$, so $\sqrt{1.2321} = 1.11$.
$1.11 = 1 + \frac{R}{100}$
$1.11 - 1 = \frac{R}{100}$
$0.11 = \frac{R}{100}$
$R = 0.11 \times 100$
$R = 11$
The calculated annual rate of interest is 11%.
If the interest earned during the $2^{nd}$ year on a certain sum is ₹3,342, and the rate of interest is 20% per annum compounded annually, then the sum is: