The question asks for the original principal sum (sum) based on the interest earned during the second year of compounding and the annual interest rate.
For annual compounding, the interest earned specifically in the nth year is given by:
Interest in nth year = $ P \times \left(1 + \frac{R}{100}\right)^{n-1} \times \frac{R}{100} $
Here, $P$ is the principal, $R$ is the rate, and $n$ is the year.
Substitute the values for the 2nd year ($n=2$):
₹6,258 = $ P \times \left(1 + \frac{20}{100}\right)^{2-1} \times \frac{20}{100} $
Simplify the equation:
₹6,258 = $ P \times \left(1 + 0.20\right)^{1} \times 0.20 $
₹6,258 = $ P \times (1.20) \times 0.20 $
₹6,258 = $ P \times 0.24 $
Solve for the principal sum ($P$):
$ P = \frac{₹6,258}{0.24} $
Perform the division:
$ P = \frac{625800}{24} = ₹26,075 $
Therefore, the principal sum is ₹26,075.
Find the interest (in ₹) on ₹8,000 at 10% per annum compounded half yearly for $1\frac{1}{2}$ years.
Amit had invested same amount of sums at simple as well as compound interest, compounded annually. The time period of investment for both the sums was 2 years and rate of interest too was the same, 4% per annum. At the end, he found a difference of ₹43 in both the interests received. What were the sums (in ₹) invested?
When the difference between compound interest, compounded annually, and simple interest for three years is ₹186 at 10% interest per annum, the principal is ₹______.