In how many years will a sum of Rs. 18,000 at 20% per annum compounded half yearly becomes Rs. 23,958?
Solving the Compound Interest Time Problem
This problem asks us to find the time period required for a principal amount to grow to a specific future value under compound interest, where the interest is compounded half-yearly.
Understanding the Given Information
We are provided with the following details:
- Principal amount (P): Rs. 18,000
- Final amount (A): Rs. 23,958
- Annual interest rate (R): 20% per annum
- Compounding frequency: Half-yearly
We need to find the number of years (n).
Compound Interest Formula for Half-Yearly Compounding
The formula for the amount (A) accumulated after 'n' years with principal (P) at an annual interest rate (R) compounded half-yearly is given by:
\(A = P \left(1 + \frac{R/2}{100}\right)^{2n}\)
Alternatively, we can think of the rate per compounding period and the total number of compounding periods.
- Rate per half-year (r) = \( \frac{R}{2} = \frac{20\%}{2} = 10\% \)
- Number of compounding periods in 'n' years = \( 2n \)
So the formula becomes:
\(A = P \left(1 + \frac{r}{100}\right)^{2n}\) or \(A = P \left(1 + \frac{10}{100}\right)^{2n}\)
\(A = P \left(1 + \frac{1}{10}\right)^{2n}\)
Step-by-Step Calculation
Now, let's substitute the given values into the formula:
\(23958 = 18000 \left(1 + \frac{1}{10}\right)^{2n}\)
\(23958 = 18000 \left(\frac{11}{10}\right)^{2n}\)
Divide both sides by the principal amount, 18000:
\(\frac{23958}{18000} = \left(\frac{11}{10}\right)^{2n}\)
Simplify the fraction on the left side:
\(\frac{23958}{18000}\)
Divide both numerator and denominator by common factors. Let's start by dividing by 6:
\(\frac{23958 \div 6}{18000 \div 6} = \frac{3993}{3000}\)
Now divide by 3:
\(\frac{3993 \div 3}{3000 \div 3} = \frac{1331}{1000}\)
So the equation becomes:
\(\frac{1331}{1000} = \left(\frac{11}{10}\right)^{2n}\)
Recognize that 1331 is the cube of 11 (\(11^3\)) and 1000 is the cube of 10 (\(10^3\)).
Therefore, \(\frac{1331}{1000} = \left(\frac{11}{10}\right)^3\)
Substitute this back into the equation:
\(\left(\frac{11}{10}\right)^3 = \left(\frac{11}{10}\right)^{2n}\)
Since the bases are the same, we can equate the exponents:
\(3 = 2n\)
Solve for n:
\(n = \frac{3}{2}\) years
\(n = 1\frac{1}{2}\) years
Conclusion
The time required for the sum of Rs. 18,000 to become Rs. 23,958 at 20% per annum compounded half-yearly is \(1\frac{1}{2}\) years.
| Initial Amount (P) | Rs. 18,000 |
|---|---|
| Final Amount (A) | Rs. 23,958 |
| Annual Rate (R) | 20% |
| Rate per half-year (r) | 10% |
| Equation | \(\frac{23958}{18000} = \left(\frac{11}{10}\right)^{2n}\) |
| Simplified Equation | \(\left(\frac{11}{10}\right)^3 = \left(\frac{11}{10}\right)^{2n}\) |
| Resulting Time (n) | \(1\frac{1}{2}\) years |
Revision Table: Compound Interest Concepts
| Term | Description | Formula Snippet |
|---|---|---|
| Principal (P) | The initial amount of money invested or borrowed. | \(P\) |
| Amount (A) | The total sum including the principal and accumulated interest after a certain period. | \(A\) |
| Interest Rate (R) | The percentage at which interest is calculated, usually per annum. | \(R\%\) |
| Time Period (n) | The duration for which the money is invested or borrowed, usually in years. | \(n\) |
| Compounding Frequency | How often the interest is added to the principal (e.g., annually, half-yearly, quarterly). | Determines the formula exponent and rate division. |
| Compound Interest (CI) | The interest earned on the principal amount as well as on the accumulated interest from previous periods. | \(CI = A - P\) |
Additional Information: Variations in Compounding
The frequency of compounding significantly affects the total interest earned. Here are common compounding frequencies and their corresponding formulas for the Amount (A) after 'n' years at an annual rate 'R%':
- Annually: Interest is added once a year.
\(A = P \left(1 + \frac{R}{100}\right)^n\)
- Half-Yearly: Interest is added twice a year.
\(A = P \left(1 + \frac{R/2}{100}\right)^{2n}\)
- Quarterly: Interest is added four times a year.
\(A = P \left(1 + \frac{R/4}{100}\right)^{4n}\)
- Monthly: Interest is added twelve times a year.
\(A = P \left(1 + \frac{R/12}{100}\right)^{12n}\)
In this problem, understanding the half-yearly compounding adjustment to both the rate and the time exponent was crucial for finding the correct duration.
Important Questions from Interest
Three years ago, the value of a flat was Rs. 65,00,000. Its value depreciated at the rate of 5%, 4% and 3% at the end of the first, the second and the third year, respectively. What is its present value?
An electric bulb was bought at Rs. 4200 Its value depreciates at the rate of 8% per annum Its value after one year will be:
What is the amount of money invested after 4 years at the rate of simple interest rate of 13% per annum invested at 4,950 rupees. (In rupees)
A certain sum of money amounts to \(\frac{3}{2}\) of itself in 2 years applying simple interest. Find the rate of simple interest per annum.
A sum of Rs. 2000 will become Rs. 2400 in 12 months at some rate of simple interest. Find the rate of interest per annum.
