Understanding the Debt and Payment Calculation

This problem involves calculating the equal annual payment required to settle a debt over a specific period, considering simple interest. The key details are:

• Principal Debt Amount ($D$): ₹15,600
• Time Period ($n$): 3 years
• Simple Interest Rate ($R$): 4% per annum (or 0.04)
• Payment Frequency: Annual
• Payment Type: Equal instalments

The goal is to find the amount of each equal annual payment ($x$) that will completely discharge the debt, including the interest accrued.

Calculating the Total Amount Due

First, we need to determine the total amount that the debtor must pay back after 3 years. This includes the original principal debt plus the simple interest accumulated over the 3 years.

The formula for Simple Interest (SI) is:

$$ SI = D \times R \times n $$

Substituting the given values:

$$ SI = 15600 \times 0.04 \times 3 $$

$$ SI = 15600 \times 0.12 $$

$$ SI = ₹1,872 $$

The total amount due ($A$) at the end of 3 years is the sum of the principal debt and the simple interest:

$$ A = D + SI $$

$$ A = 15600 + 1872 $$

$$ A = ₹17,472 $$

This total amount of ₹17,472 must be paid off through equal annual instalments over the 3 years.

Determining the Equal Annual Payment (Standard Method)

When paying a debt in equal instalments under simple interest, the sum of the future values of all instalments at the end of the term must equal the total amount due. Let the equal annual payment be $x$. Assuming the payments are made at the end of each year:

• The first payment (made at the end of year 1) earns simple interest for the remaining 2 years ($n-1 = 2$). Its future value at the end of year 3 is $x(1 + R \times (n-1)) = x(1 + 0.04 \times 2) = x(1.08)$.
• The second payment (made at the end of year 2) earns simple interest for the remaining 1 year ($n-2 = 1$). Its future value at the end of year 3 is $x(1 + R \times (n-2)) = x(1 + 0.04 \times 1) = x(1.04)$.
• The third payment (made at the end of year 3) earns simple interest for 0 years ($n-3 = 0$). Its future value at the end of year 3 is $x(1 + R \times (n-3)) = x(1 + 0.04 \times 0) = x$.

The sum of the future values of these instalments must equal the total amount due (₹17,472).

$$ x(1 + (n-1)R) + x(1 + (n-2)R) + ... + x = A $$

For $n=3$:

$$ x(1 + 2R) + x(1 + R) + x = A $$

$$ x(1 + 2(0.04)) + x(1 + 0.04) + x = 17472 $$

$$ x(1 + 0.08) + x(1.04) + x = 17472 $$

$$ 1.08x + 1.04x + x = 17472 $$

$$ (1.08 + 1.04 + 1)x = 17472 $$

$$ 3.12x = 17472 $$

Now, solve for $x$:

$$ x = \frac{17472}{3.12} $$

$$ x = 5600 $$

Therefore, according to this standard calculation method, the equal annual payment should be ₹5,600.

Conclusion and Provided Answer Analysis

The calculation using the standard method for simple interest annuities indicates that the equal annual payment required is ₹5,600. However, the provided correct answer is ₹5,000. There seems to be a discrepancy between the calculated value based on standard financial mathematics principles and the provided answer. This could potentially stem from a non-standard interpretation of simple interest application in annuities, a simplification, or an error in the question's parameters or the provided answer key. Based on the standard approach, ₹5,600 is the derived amount.