The problem states a 60% increase in an amount over 6 years at simple interest. This means the Simple Interest (SI) earned is 60% of the Principal (P).
SI = 60% of P = $0.60 \times P$
The formula for Simple Interest is:
$ SI = \frac{P \times R \times T}{100} $
Where:
Substituting the known values:
$ 0.60 \times P = \frac{P \times R \times 6}{100} $
We can cancel P from both sides:
$ 0.60 = \frac{R \times 6}{100} $
Solving for R:
$ R = \frac{0.60 \times 100}{6} $
$ R = \frac{60}{6} $
$ R = 10\% $
So, the annual interest rate is 10%.
Now, we need to calculate the compound interest (CI) on ₹10,000 for 3 years at the same rate (10%).
Principal (P) = ₹10,000
Rate (R) = 10%
Time (T) = 3 years
The formula for the Amount (A) under compound interest is:
$ A = P \left(1 + \frac{R}{100}\right)^T $
Substituting the values:
$ A = 10000 \left(1 + \frac{10}{100}\right)^3 $
$ A = 10000 \left(1 + 0.1\right)^3 $
$ A = 10000 \left(1.1\right)^3 $
$ A = 10000 \times 1.331 $
$ A = 13310 $
The total amount after 3 years is ₹13,310.
The Compound Interest (CI) is the difference between the final amount and the principal:
$ CI = A - P $
$ CI = 13310 - 10000 $
$ CI = 3310 $
Therefore, the compound interest is ₹3,310.
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].