This problem requires us to find the annual rate of interest ('R') given the principal amount ('P'), the final amount ('A'), the time period ('t'), and the compounding frequency (half-yearly).
We are given the following information:
We need to find the Rate of Interest (R) per annum.
Since the interest is compounded half-yearly, we need to adjust the time period and the rate of interest accordingly.
The formula for the final amount (A) with compound interest is:
$$ A = P \left(1 + \frac{r}{100}\right)^n $$
Substituting the rate per period (R/2) into the formula, we get:
$$ A = P \left(1 + \frac{R}{2 \times 100}\right)^n $$
$$ A = P \left(1 + \frac{R}{200}\right)^n $$
Now, let's plug in the given values:
$$ 15625 = 8000 \left(1 + \frac{R}{200}\right)^3 $$
To solve for R, we first isolate the term containing R:
$$ \frac{15625}{8000} = \left(1 + \frac{R}{200}\right)^3 $$
Let's simplify the fraction $\frac{15625}{8000}$:
So, the equation becomes:
$$ \frac{125}{64} = \left(1 + \frac{R}{200}\right)^3 $$
We need to find the cube root of both sides. We know that $5^3 = 125$ and $4^3 = 64$. Therefore:
$$ \left(\frac{5}{4}\right)^3 = \left(1 + \frac{R}{200}\right)^3 $$
Now, we can equate the bases:
$$ \frac{5}{4} = 1 + \frac{R}{200} $$
Let's solve for R:
The rate of interest is 50% per annum.
The certain sum amounts to Rs. 9,982.50 in \(2\frac{1}{2}\) years at 12% p.a., interest compounded 10-monthly. The sum (in Rs.) is:
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If a sum of ₹ 2000 is lent at 10% p.a. compound interest, what is the interest for the second year?
A sum becomes 5 times of itself in 3 years. at compound interest (interest is compounded annually). In how many years. will the sum becomes 125 times of itself?
If the compound interest on a certain sum of money for two years at 9% p.a. is Rs. 3,762, then the sum is: