The rate of interest on two different schemes is the same and it is 20%. But in one of the schemes, the interest is compounded half-yearly and in the other, the interest is compounded annually. Equal amounts are invested in the schemes. If the difference of the returns after 2 years is Rs. 482, then what is the principal amount in each scheme?
Rs. 20,000
Understanding the Problem: Compound Interest Schemes
This question involves comparing two investment schemes that offer the same nominal rate of interest but differ in their compounding frequency. One scheme compounds interest half-yearly, while the other compounds annually. We are given that equal amounts are invested in both schemes for 2 years, and the difference in the compound interest earned is Rs. 482. Our goal is to find the principal amount invested in each scheme.
Key Concepts: Compound Interest Calculation
The formula for the amount (A) under compound interest is given by:
\( A = P(1 + \frac{r}{n})^{nt} \)
Where:
- \( P \) is the principal amount.
- \( r \) is the annual interest rate (as a decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time the money is invested for in years.
The compound interest (CI) is the amount minus the principal:
\( CI = A - P \)
Scheme 1: Interest Compounded Half-Yearly
In this scheme, the interest is compounded twice a year (n = 2). The annual rate is 20% (r = 0.20). The time period is 2 years (t = 2). The principal amount is the same for both schemes, let's call it P.
The rate per compounding period is \( \frac{r}{n} = \frac{0.20}{2} = 0.10 \) (or 10%).
The total number of compounding periods is \( nt = 2 \times 2 = 4 \).
The amount after 2 years in Scheme 1 (\(A_1\)) is:
\( A_1 = P(1 + 0.10)^{4} \)
\( A_1 = P(1.1)^{4} \)
Calculating \( (1.1)^4 \):
\( (1.1)^2 = 1.21 \)
\( (1.1)^4 = (1.21)^2 = 1.4641 \)
So, \( A_1 = 1.4641P \)
The compound interest from Scheme 1 (\(CI_1\)) is:
\( CI_1 = A_1 - P = 1.4641P - P = 0.4641P \)
Scheme 2: Interest Compounded Annually
In this scheme, the interest is compounded once a year (n = 1). The annual rate is 20% (r = 0.20). The time period is 2 years (t = 2). The principal amount is P.
The rate per compounding period is \( \frac{r}{n} = \frac{0.20}{1} = 0.20 \) (or 20%).
The total number of compounding periods is \( nt = 1 \times 2 = 2 \).
The amount after 2 years in Scheme 2 (\(A_2\)) is:
\( A_2 = P(1 + 0.20)^{2} \)
\( A_2 = P(1.2)^{2} \)
Calculating \( (1.2)^2 \):
\( (1.2)^2 = 1.44 \)
So, \( A_2 = 1.44P \)
The compound interest from Scheme 2 (\(CI_2\)) is:
\( CI_2 = A_2 - P = 1.44P - P = 0.44P \)
Finding the Principal Amount (P)
The difference between the returns (compound interest) from the two schemes is given as Rs. 482.
Difference = \( CI_1 - CI_2 \)
\( 482 = 0.4641P - 0.44P \)
\( 482 = (0.4641 - 0.44)P \)
\( 482 = 0.0241P \)
Now, we need to solve for P:
\( P = \frac{482}{0.0241} \)
To simplify the division, multiply the numerator and denominator by 10000 to remove the decimal:
\( P = \frac{482 \times 10000}{0.0241 \times 10000} = \frac{4820000}{241} \)
Let's perform the division:
\( \frac{4820000}{241} = \frac{482 \times 10000}{241} \)
Since \( 482 = 2 \times 241 \):
\( P = \frac{(2 \times 241) \times 10000}{241} \)
\( P = 2 \times 10000 \)
\( P = 20000 \)
The principal amount invested in each scheme is Rs. 20,000.
Summary of Compound Interest Calculation
| Scheme | Compounding Frequency (n) | Rate per Period (\(r/n\)) | Number of Periods (nt) | Amount Formula (\(A\)) | CI Formula (\(CI\)) | Calculated CI |
|---|---|---|---|---|---|---|
| Scheme 1 (Half-yearly) | 2 | \(20\%/2 = 10\% = 0.1\) | \(2 \text{ years} \times 2 = 4\) | \(P(1+0.1)^4\) | \(P(1.1)^4 - P\) | \(1.4641P - P = 0.4641P\) |
| Scheme 2 (Annually) | 1 | \(20\%/1 = 20\% = 0.2\) | \(2 \text{ years} \times 1 = 2\) | \(P(1+0.2)^2\) | \(P(1.2)^2 - P\) | \(1.44P - P = 0.44P\) |
Difference in CI: \( 0.4641P - 0.44P = 0.0241P \)
Given difference: Rs. 482
\( 0.0241P = 482 \)
\( P = \frac{482}{0.0241} = 20000 \)
The principal amount is Rs. 20,000.
Revision Table: Compound Interest Concepts
| Concept | Description | Formula |
|---|---|---|
| Compound Interest (CI) | Interest calculated on the initial principal and also on the accumulated interest of previous periods. | \(CI = A - P\) |
| Amount (A) | The total sum received at the end of the investment period, including principal and interest. | \(A = P(1 + \frac{r}{n})^{nt}\) |
| Nominal Annual Rate (r) | The stated annual interest rate, before accounting for compounding frequency. | Given in the problem (20% here). |
| Rate per Period (\(r/n\)) | The interest rate applied during each compounding interval. | \(r\) divided by the number of compounding periods per year. |
| Compounding Frequency (n) | How many times interest is calculated and added to the principal within one year. | 1 for annually, 2 for half-yearly, 4 for quarterly, 12 for monthly. |
Additional Information: Effective Annual Rate (EAR)
When interest is compounded more frequently than annually, the actual rate of return over a year is higher than the nominal annual rate. This is called the Effective Annual Rate (EAR).
The formula for EAR is:
\( \text{EAR} = (1 + \frac{r}{n})^{n} - 1 \)
For Scheme 1 (Half-yearly, r=0.20, n=2):
\( \text{EAR}_1 = (1 + \frac{0.20}{2})^{2} - 1 = (1 + 0.10)^{2} - 1 = (1.1)^2 - 1 = 1.21 - 1 = 0.21 \)
This means Scheme 1 effectively yields 21% per year.
For Scheme 2 (Annually, r=0.20, n=1):
\( \text{EAR}_2 = (1 + \frac{0.20}{1})^{1} - 1 = (1.2)^{1} - 1 = 1.20 - 1 = 0.20 \)
This means Scheme 2 effectively yields 20% per year.
The difference in returns over 2 years comes from the difference in these effective rates compounded over the 2-year period.
Using EAR to find the amount after 2 years:
\( A_1 = P(1 + \text{EAR}_1)^2 = P(1 + 0.21)^2 = P(1.21)^2 = 1.4641P \)
\( A_2 = P(1 + \text{EAR}_2)^2 = P(1 + 0.20)^2 = P(1.20)^2 = 1.44P \)
These amounts match the amounts calculated earlier using the per-period rate and number of periods, confirming the consistency of the compound interest formulas.
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