Find the compound interest on ₹42000 for 1½ years at 10% p.a. compounded annually.
₹6510
The question asks us to find the compound interest on a principal amount for a time period that is not a whole number of years (1½ years). When calculating compound interest for a period like 1½ years compounded annually, we calculate the amount for the whole number of years (1 year) and then calculate simple interest on this amount for the remaining fractional part of the year (½ year).
First, we calculate the amount after 1 full year using the compound interest formula:
Amount (A) = \( P \left(1 + \frac{R}{100}\right)^n \)
Here, P = ₹42000, R = 10%, and n = 1 year.
Amount after 1 year \( A_1 = 42000 \left(1 + \frac{10}{100}\right)^1 \)
\( A_1 = 42000 \left(1 + 0.1\right) \)
\( A_1 = 42000 \times 1.1 \)
\( A_1 = ₹46200 \)
So, the amount after 1 year is ₹46200.
Now, this amount \( A_1 \) becomes the principal for the next ½ year. We calculate simple interest on this amount for the remaining time period (½ year) at the given annual rate (10%).
Principal for simple interest \( P' = A_1 = ₹46200 \)
Rate \( R' = 10\% \) p.a.
Time \( t = \frac{1}{2} \) years \( = 0.5 \) years
Simple Interest (SI) = \( \frac{P' \times R' \times t}{100} \)
\( SI = \frac{46200 \times 10 \times 0.5}{100} \)
\( SI = \frac{46200 \times 5}{100} \)
\( SI = 462 \times 5 \)
\( SI = ₹2310 \)
The simple interest for the remaining ½ year is ₹2310.
The total amount after 1½ years is the amount after 1 year plus the simple interest for the next ½ year.
Total Amount \( A = A_1 + SI \)
\( A = 46200 + 2310 \)
\( A = ₹48510 \)
The total amount after 1½ years is ₹48510.
Compound Interest (CI) is the difference between the total amount and the original principal.
Compound Interest = Total Amount - Original Principal
\( CI = A - P \)
\( CI = 48510 - 42000 \)
\( CI = ₹6510 \)
The compound interest on ₹42000 for 1½ years at 10% p.a. compounded annually is ₹6510.
| Calculation Step | Formula/Method | Value |
|---|---|---|
| Principal (P) | Given | ₹42000 |
| Rate (R) | Given | 10% p.a. |
| Time (n) | Given | 1½ years |
| Amount after 1 year (\( A_1 \)) | \( P \left(1 + \frac{R}{100}\right)^1 \) | \( 42000 \times 1.1 = ₹46200 \) |
| Simple Interest for ½ year (SI) | \( \frac{A_1 \times R \times \text{fractional time}}{100} \) | \( \frac{46200 \times 10 \times 0.5}{100} = ₹2310 \) |
| Total Amount (A) | \( A_1 + SI \) | \( 46200 + 2310 = ₹48510 \) |
| Compound Interest (CI) | \( A - P \) | \( 48510 - 42000 = ₹6510 \) |
Let's compare our calculated compound interest with the given options:
Our calculated compound interest is ₹6510, which matches Option 3.
To find the compound interest for a period with a fraction of a year, we treat the whole number of years as compounded annually and the remaining fraction of a year as simple interest on the amount accumulated at the end of the whole years. This approach correctly handles the compounding effect for the full periods and applies the annual rate proportionally for the final partial period.
| Concept | Description | Formula (for annual compounding) |
|---|---|---|
| Principal (P) | The initial amount of money invested or borrowed. | N/A |
| Rate (R) | The annual interest rate. | N/A |
| Time (n) | The duration for which the money is invested or borrowed. | N/A |
| Amount (A) | The total sum of the principal and interest after a given time. | \( A = P \left(1 + \frac{R}{100}\right)^n \) (for whole years) |
| Compound Interest (CI) | The interest earned on the principal and on the accumulated interest from previous periods. | \( CI = A - P \) |
| Simple Interest (SI) | Interest calculated only on the principal amount. | \( SI = \frac{P \times R \times T}{100} \) |
When calculating compound interest for a period of \( n \) whole years and a fraction \( f \) of a year (like \( n + f \)), the formula used implicitly is:
\( \text{Amount} = P \left(1 + \frac{R}{100}\right)^n \left(1 + \frac{f \times R}{100}\right) \)
Let's test this formula with our values: P = 42000, R = 10, n = 1, f = 0.5.
\( \text{Amount} = 42000 \left(1 + \frac{10}{100}\right)^1 \left(1 + \frac{0.5 \times 10}{100}\right) \)
\( \text{Amount} = 42000 \left(1.1\right) \left(1 + \frac{5}{100}\right) \)
\( \text{Amount} = 42000 \times 1.1 \times (1 + 0.05) \)
\( \text{Amount} = 42000 \times 1.1 \times 1.05 \)
\( \text{Amount} = 46200 \times 1.05 \)
\( \text{Amount} = ₹48510 \)
This formula gives the total amount, which matches our step-by-step calculation. The compound interest is then \( 48510 - 42000 = ₹6510 \). This confirms that our step-by-step method is equivalent to using this combined formula for fractional years when compounding is annual.
It's important to note that this method for fractional years applies when compounding is done annually. If compounding were done half-yearly or quarterly, the approach for fractional periods within those compounding cycles would differ.
Simple interest on a certain sum at 8% per annum for 5 years is ₹2400. What will be the compound interest on the same sum at the same rate for 2 years?
Find the compound interest on ₹64000 at 10% per annum for 9 months when interest is compounded quarterly.
A man invested on simple interest, 1/4 of his capital at 7% p.a., another 1/4 of the capital at 8% p.a. and the remaining capital at 10% p.a. He earned Rs. 700 as interest in one year. Find his total capital invested.
A sum of money doubles itself in 10 years on compound interest. In how many years will it become four times?
A sum of ₹1500 is invested at 5% p.a. simple interest. If the interest is added to the principal after every 10 years, the amount will become ₹3000 after: