This problem involves calculating the annual rate of interest when interest is compounded half-yearly.
The formula for compound interest is:
$$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$
Where:
In this case, the interest is compounded half-yearly. This means:
The formula becomes:
$$ A = P \left(1 + \frac{r/2}{100}\right)^{3} $$
Or simplified:
$$ A = P \left(1 + \frac{r}{200}\right)^{3} $$
We are given:
Substitute these values into the formula:
$$ 19,683 = 15,625 \left(1 + \frac{r}{200}\right)^{3} $$
Divide both sides by the principal amount (₹15,625):
$$ \frac{19,683}{15,625} = \left(1 + \frac{r}{200}\right)^{3} $$
To solve for the term inside the parenthesis, we need to find the cube root of both sides:
$$ \sqrt[3]{\frac{19,683}{15,625}} = 1 + \frac{r}{200} $$
We need to find the cube roots of 19,683 and 15,625.
Let's check potential cubes:
So, the cube root is:
$$ \frac{27}{25} = 1 + \frac{r}{200} $$
Now, isolate the rate ($r$):
$$ \frac{r}{200} = \frac{27}{25} - 1 $$
Find a common denominator:
$$ \frac{r}{200} = \frac{27}{25} - \frac{25}{25} $$
$$ \frac{r}{200} = \frac{2}{25} $$
Multiply both sides by 200:
$$ r = \frac{2}{25} \times 200 $$
$$ r = 2 \times \frac{200}{25} $$
$$ r = 2 \times 8 $$
$$ r = 16 $$
The annual rate of interest ($r$) is 16%.
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:
The difference between the simple interest and the compound interest compounded annually on a certain sum of money for 2 years at a rate of 8% per annum is Rs. 16.80. Find the principle amount.
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: