At what rate of compound interest per annum, a sum of Rs. 20000 becomes Rs. 23152.50 in 1 year 6 months compounded half-yearly?
10% per annum
The question asks us to find the annual rate of compound interest at which a given principal amount grows to a specific amount over a certain period, with interest compounded half-yearly.
Let's identify the key information provided:
When interest is compounded half-yearly, the number of compounding periods per year (n) is 2. The total number of compounding periods over the given time will be $n \times t$. The rate of interest applied per compounding period will be the annual rate divided by n, i.e., $\frac{r}{n}$, where r is the annual interest rate.
The formula for compound interest is:
$$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$
Now, we substitute the given values into the formula:
$$ 23152.50 = 20000 \left(1 + \frac{r}{2}\right)^{2 \times 1.5} $$
$$ 23152.50 = 20000 \left(1 + \frac{r}{2}\right)^{3} $$
To find the rate (r), we need to isolate the term containing r. First, divide both sides of the equation by the principal amount, 20000:
$$ \frac{23152.50}{20000} = \left(1 + \frac{r}{2}\right)^{3} $$
Calculate the value on the left side:
$$ 1.157625 = \left(1 + \frac{r}{2}\right)^{3} $$
Now, take the cube root of both sides to eliminate the exponent of 3:
$$ \sqrt[3]{1.157625} = 1 + \frac{r}{2} $$
The cube root of 1.157625 is 1.05:
$$ 1.05 = 1 + \frac{r}{2} $$
Next, subtract 1 from both sides:
$$ 1.05 - 1 = \frac{r}{2} $$
$$ 0.05 = \frac{r}{2} $$
Finally, multiply both sides by 2 to find the value of r:
$$ r = 0.05 \times 2 $$
$$ r = 0.10 $$
The value of r is 0.10. Since r is the annual rate expressed as a decimal, we convert it to a percentage by multiplying by 100:
$$ \text{Annual Rate} = 0.10 \times 100\% = 10\% $$
So, the compound interest rate per annum is 10%.
Let's summarize the calculation steps:
| Step | Description | Calculation |
|---|---|---|
| 1 | Set up the compound interest formula | $A = P \left(1 + \frac{r}{n}\right)^{nt}$ |
| 2 | Substitute known values | $23152.50 = 20000 \left(1 + \frac{r}{2}\right)^{3}$ |
| 3 | Isolate the term with r | $\frac{23152.50}{20000} = \left(1 + \frac{r}{2}\right)^{3}$ → $1.157625 = \left(1 + \frac{r}{2}\right)^{3}$ |
| 4 | Take the cube root | $\sqrt[3]{1.157625} = 1 + \frac{r}{2}$ → $1.05 = 1 + \frac{r}{2}$ |
| 5 | Solve for $\frac{r}{2}$ | $1.05 - 1 = \frac{r}{2}$ → $0.05 = \frac{r}{2}$ |
| 6 | Solve for r (annual rate) | $r = 0.05 \times 2$ → $r = 0.10$ |
| 7 | Convert to percentage | $0.10 \times 100\%$ → $10\%$ |
The calculated annual compound interest rate is 10%.
| Concept | Explanation | Formula Snippet |
|---|---|---|
| Principal (P) | The initial amount of money invested or borrowed. | - |
| Amount (A) | The total sum after adding interest to the principal. | $A = P + \text{Interest}$ |
| Rate (r) | The annual rate of interest, usually expressed as a decimal or percentage. | - |
| Time (t) | The duration for which the money is invested or borrowed, in years. | - |
| Compounding Frequency (n) | The number of times interest is calculated and added to the principal per year (e.g., 1 for annually, 2 for half-yearly, 4 for quarterly, 12 for monthly). | - |
| Compound Interest (CI) | Interest calculated on the initial principal and also on the accumulated interest of previous periods. | $CI = A - P$ |
The compounding frequency significantly impacts the total amount earned or paid. More frequent compounding leads to higher interest over the same period and rate because interest starts earning interest sooner.
Here are common compounding frequencies and their 'n' values:
In this problem, 'half-yearly' compounding was key, meaning we used n=2 and adjusted both the rate per period ($\frac{r}{2}$) and the total number of periods ($nt = 1.5 \times 2 = 3$). Understanding the compounding period is crucial for solving compound interest problems correctly.
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