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:
7,500
Understanding the Compound Interest Problem
This question asks us to find the original sum of money (principal) that grows to a specific amount when compounded at a certain rate for a given time period. The key detail here is that the interest is compounded 10-monthly, which is not the standard annual, semi-annual, or quarterly compounding.
Given Information
- Amount (A) = Rs. 9,982.50
- Time (t) = \(2\frac{1}{2}\) years = 2.5 years
- Nominal Annual Interest Rate (R) = 12% p.a.
- Compounding Frequency: 10-monthly
Calculating the Effective Rate and Number of Periods
Since the interest is compounded 10-monthly, we need to adjust the annual rate and the total time into terms of 10-month periods.
Rate per Compounding Period
An year has 12 months. The interest is compounded every 10 months. So, the number of compounding periods in a year is \(12 / 10 = 1.2\).
The interest rate per compounding period (i) is the annual rate divided by the number of periods per year:
\(i = \frac{\text{Nominal Annual Rate}}{\text{Number of Compounding Periods per Year}}\)
\(i = \frac{12\%}{1.2} = \frac{0.12}{1.2} = 0.10\)
So, the interest rate per 10-month period is 10% or 0.10.
Total Number of Compounding Periods
The total time is 2.5 years. We need to find out how many 10-month periods are there in 2.5 years.
Total time in months = \(2.5 \text{ years} \times 12 \text{ months/year} = 30 \text{ months}\)
Number of compounding periods (N) = \(\frac{\text{Total Time in Months}}{\text{Duration of one compounding period}}\)
\(N = \frac{30 \text{ months}}{10 \text{ months/period}} = 3 \text{ periods}\)
Applying the Compound Interest Formula
The formula for the amount (A) in compound interest is:
\(A = P(1 + i)^N\)
Where:
- P is the principal sum
- i is the interest rate per compounding period
- N is the total number of compounding periods
We know A, i, and N, and we need to find P. We can rearrange the formula to solve for P:
\(P = \frac{A}{(1 + i)^N}\)
Step-by-Step Calculation
Substitute the values we found into the formula for P:
\(P = \frac{9982.50}{(1 + 0.10)^3}\)
\(P = \frac{9982.50}{(1.10)^3}\)
Calculate \((1.10)^3\):
\((1.10)^3 = 1.10 \times 1.10 \times 1.10 = 1.21 \times 1.10 = 1.331\)
Now substitute this back into the equation for P:
\(P = \frac{9982.50}{1.331}\)
Perform the division:
\(P = 7500\)
Conclusion
The original sum (principal) is Rs. 7,500.
| Parameter | Value | Calculation/Explanation |
|---|---|---|
| Amount (A) | Rs. 9982.50 | Given |
| Time (t) | 2.5 years | Given (\(2\frac{1}{2}\) years) |
| Annual Rate (R) | 12% p.a. | Given |
| Compounding Frequency | 10-monthly | Given |
| Rate per Period (i) | 10% or 0.10 | \(12\% / (12/10) = 12\% / 1.2\) |
| Total Periods (N) | 3 | \(2.5 \text{ years} \times (12/10) \text{ periods/year} = 2.5 \times 1.2\) or \(30 \text{ months} / 10 \text{ months/period}\) |
| Principal (P) | Rs. 7500 | \(P = A / (1+i)^N\) |
Revision Table: Compound Interest Basics
| Concept | Description | Formula (for annual compounding) |
|---|---|---|
| Principal (P) | The initial sum of money. | - |
| Amount (A) | The total sum after adding interest to the principal. | \(A = P(1 + r/100)^t\) |
| Interest Rate (r) | The rate at which interest is calculated (usually annual). | - |
| Time (t) | The duration for which the money is invested or borrowed. | - |
| Compound Interest (CI) | Interest calculated on the principal amount and also on the accumulated interest of previous periods. | \(CI = A - P\) or \(CI = P[(1 + r/100)^t - 1]\) |
| Compounding Frequency | How often interest is added to the principal (e.g., annually, semi-annually, quarterly, monthly). | Affects the value of 'i' and 'N' in the general formula \(A = P(1 + i)^N\). |
Additional Information: Handling Different Compounding Periods
When interest is compounded more than once a year, we use the general formula \(A = P(1 + i)^N\). Here's how to find 'i' and 'N':
- Let R be the annual interest rate (as a decimal).
- Let t be the time in years.
- Let n be the number of times interest is compounded per year.
- The rate per compounding period is \(i = R/n\).
- The total number of compounding periods is \(N = t \times n\).
Examples of 'n':
- Annually: n = 1
- Semi-annually: n = 2
- Quarterly: n = 4
- Monthly: n = 12
- Weekly: n = 52
- Daily: n = 365 (usually)
In our problem, the compounding is 10-monthly. To find 'n', we ask how many 10-month periods are there in 12 months (one year). \(n = 12 / 10 = 1.2\). This is a less common frequency, but the principle remains the same. Once 'n' is found, 'i' and 'N' are calculated as shown in the solution above.
Important Questions from Compound Interest
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:
The compound interest on a sum of ₹ 24500 at 10% p.a for \(2\frac{2}{5}\) years interest compounded yearly is:
