The formula for compound interest (CI) is:
$ CI = P \left[ \left(1 + \frac{R}{100}\right)^T - 1 \right] $
We are given:
Substitute the values into the formula to find the Principal (P):
$ 6305 = P \left[ \left(1 + \frac{5}{100}\right)^3 - 1 \right] $
$ 6305 = P \left[ \left(1 + 0.05\right)^3 - 1 \right] $
$ 6305 = P \left[ (1.05)^3 - 1 \right] $
Calculate $(1.05)^3$:
$ (1.05)^3 = 1.157625 $
Now, substitute this back:
$ 6305 = P \left[ 1.157625 - 1 \right] $
$ 6305 = P \left[ 0.157625 \right] $
Solve for P:
$ P = \frac{6305}{0.157625} $
$ P = 40000 $
So, the principal sum is ₹40,000.
Now, we need to find the simple interest (SI) for the same principal, rate, and time.
The formula for simple interest is:
$ SI = \frac{P \times R \times T}{100} $
Using the calculated principal (P = ₹40,000), R = 5%, and T = 3 years:
$ SI = \frac{40000 \times 5 \times 3}{100} $
$ SI = 400 \times 5 \times 3 $
$ SI = 2000 \times 3 $
$ SI = 6000 $
The simple interest is ₹6,000.
Find the interest (in ₹) on ₹8,000 at 10% per annum compounded half yearly for $1\frac{1}{2}$ years.
The difference between the simple interest and the compound interest, compounded annually, on a certain sum of money for 2 years at 17% per annum is ₹967. Find the sum [rounded off to the nearest integer].