A sum of Rs. P was borrowed and paid back in two equal yearly instalments, each of Rs. 35,280. If the rate of interest was 5% per annum and interest is compounding annually, then the value of P is ________.
Rs. 65,600
Calculating Loan Principal with Equal Yearly Installments
This problem involves a loan that is repaid through a series of equal yearly installments. The key concept here is understanding the present value of these future payments when the interest is compounded annually. The original sum borrowed (P) is equal to the sum of the present values of all the installments.
We are given the following information:
- Each yearly installment amount = Rs. 35,280
- Number of installments = 2
- Rate of interest (r) = 5% per annum
- Interest compounding frequency = Annually
The rate of interest can be written as a decimal:
\(r = 5\% = \frac{5}{100} = 0.05\)
The formula to calculate the present value (PV) of a single amount (A) received after \(n\) years with a compound annual interest rate \(r\) is:
$$PV = \frac{A}{(1 + r)^n}$$
In this problem, the principal amount P is the sum of the present values of the two installments. The first installment is paid after 1 year, and the second after 2 years.
Step-by-Step Calculation of Present Values
Present Value of the First Installment
The first installment of Rs. 35,280 is paid after 1 year (\(n=1\)). Its present value is:
$$PV_1 = \frac{35280}{(1 + 0.05)^1} = \frac{35280}{1.05}$$
Calculating the value:
\(PV_1 = 33600\)
The present value of the first installment is Rs. 33,600.
Present Value of the Second Installment
The second installment of Rs. 35,280 is paid after 2 years (\(n=2\)). Its present value is:
$$PV_2 = \frac{35280}{(1 + 0.05)^2} = \frac{35280}{(1.05)^2} = \frac{35280}{1.1025}$$
Calculating the value:
\(PV_2 = 32000\)
The present value of the second installment is Rs. 32,000.
Total Principal Amount (P)
The original principal amount P is the sum of the present values of all the installments.
$$P = PV_1 + PV_2$$
$$P = 33600 + 32000$$
$$P = 65600$$
Thus, the value of P is Rs. 65,600.
| Installment Number | Amount (A) | Years (n) | Present Value Calculation | Present Value |
|---|---|---|---|---|
| 1 | 35280 | 1 | \(\frac{35280}{(1.05)^1}\) | 33600 |
| 2 | 35280 | 2 | \(\frac{35280}{(1.05)^2}\) | 32000 |
Total Principal \(P = 33600 + 32000 = 65600\)
The value of P is Rs. 65,600.
Revision Table: Loan Repayment Concepts
| Concept | Description | Formula/Relevance |
|---|---|---|
| Principal (P) | The initial amount borrowed. | Sum of present values of all installments. |
| Installment (A) | The fixed amount paid periodically. | In this case, yearly equal payment. |
| Interest Rate (r) | Rate at which interest is charged. | Applied to the outstanding balance or for discounting future values. |
| Compounding Annually | Interest is added to the principal once a year. | Affects the factor \((1+r)^n\). |
| Present Value (PV) | The current value of a future sum of money given a specific rate of return. | \(PV = \frac{A}{(1 + r)^n}\) (for a single future amount). |
Additional Information: Annuity and Loan Calculations
This problem is an example of an annuity problem, specifically a loan repaid by an ordinary annuity (payments made at the end of each period). The formula for the present value of an ordinary annuity is:
$$PV_{annuity} = A \left[ \frac{1 - (1 + r)^{-n}}{r} \right]$$
Where:
- \(A\) = Installment amount
- \(r\) = Interest rate per period
- \(n\) = Total number of periods
For a loan repaid by equal installments, the principal (P) is equal to the present value of the annuity.
In our specific case with only two installments, we can verify this formula, but the method of summing individual present values is simpler and more direct for a small number of periods.
Let's quickly check with the annuity formula:
\(r = 0.05\), \(n = 2\), \(A = 35280\)
$$P = 35280 \left[ \frac{1 - (1 + 0.05)^{-2}}{0.05} \right]$$
$$P = 35280 \left[ \frac{1 - (1.05)^{-2}}{0.05} \right]$$
$$P = 35280 \left[ \frac{1 - \frac{1}{(1.05)^2}}{0.05} \right]$$
$$P = 35280 \left[ \frac{1 - \frac{1}{1.1025}}{0.05} \right]$$
$$P = 35280 \left[ \frac{\frac{1.1025 - 1}{1.1025}}{0.05} \right] = 35280 \left[ \frac{\frac{0.1025}{1.1025}}{0.05} \right]$$
$$P = 35280 \left[ \frac{0.1025}{1.1025 \times 0.05} \right] = 35280 \left[ \frac{0.1025}{0.055125} \right]$$
$$P = 35280 \times 1.863915...$$
Using precise values for \(1/1.05\) and \(1/(1.05)^2\):
\(1/1.05 \approx 0.952381\)
\(1/(1.05)^2 \approx 0.907029\)
\(P = 35280 \times 0.952381 + 35280 \times 0.907029\)
\(P = 33600 + 32000 = 65600\)
Both methods yield the same result, confirming the calculation for the loan principal P.
Important Questions from Installments
A loan of Rs. 1,50,000 is availed with compound interest rate of 10% per annum for two years compounded annually. It is to be paid in equal yearly installments, and the installment is to be paid at the end of each year. The value of the equal yearly installment is : (Rounded off to two places of decimal)
A sum of Rs. 16400 is borrowed to be paid back in 2 years by equal payments allowing 5% compound interest. Find the annual payment.
A sum of Rs. 1100 was taken as a loan. This is to be paid in two equal installments. If the rate of interest is 20% per annum, compounded annually, find the amount payable in each installment.
The formula for finding the annual installment, when A is the amount taken on loan, where r% is the rate of interest, n is the number of installments, is:
A loan of Rs 15000 is to be repaid in 4 equal annual installments. If the compound interest rate is 10% per annum, what is the approximate amount of each installment?
