Solving Loan Repayment with Equal Annual Installments

This problem involves calculating the value of equal annual installments paid to repay a loan taken at a specific compound interest rate. Rajnish borrowed a sum of money and repaid it over two years with interest compounded annually.

Understanding the Problem: Loan Installments

When a loan is repaid in equal installments, each installment consists of a portion of the principal and the interest accumulated up to that point. A common way to approach this type of problem, especially with compound interest, is to consider the present value of each installment. The sum of the present values of all future installments must equal the original principal amount borrowed.

Given Information:

• Principal amount borrowed ($\text{P}$): ₹1,500
• Rate of interest ($\text{R}$): 40% per annum, compounded annually
• Number of equal annual installments: 2
• First installment paid after 1 year, second after 2 years.

Let the value of each equal annual installment be $X$ rupees.

Calculating the Present Value of Installments

The present value of an amount $A$ received after $n$ years, compounded at a rate $R$ per annum, is given by the formula:

$\text{PV} = \frac{A}{\left(1 + \frac{R}{100}\right)^n}$

In this case, the rate of interest is 40% per annum, so $\left(1 + \frac{R}{100}\right) = \left(1 + \frac{40}{100}\right) = (1 + 0.4) = 1.4$.

The present value of the first installment, paid after 1 year, is $\frac{X}{(1.4)^1}$.

The present value of the second installment, paid after 2 years, is $\frac{X}{(1.4)^2}$.

The sum of the present values of these two installments must equal the original principal amount borrowed:

$\text{P} = \frac{X}{(1.4)^1} + \frac{X}{(1.4)^2}$

Substitute the value of $\text{P} = 1500$:

$1500 = \frac{X}{1.4} + \frac{X}{(1.4)^2}$

Solving for the Installment Value ($X$)

First, calculate $(1.4)^2$:

$(1.4)^2 = 1.4 \times 1.4 = 1.96$

Now substitute this value back into the equation:

$1500 = \frac{X}{1.4} + \frac{X}{1.96}$

To solve for $X$, we can factor out $X$ and simplify the terms:

$1500 = X \left( \frac{1}{1.4} + \frac{1}{1.96} \right)$

Find a common denominator, which is 1.96:

$1500 = X \left( \frac{1.96/1.4}{1.96} + \frac{1}{1.96} \right)$

$1500 = X \left( \frac{1.4}{1.96} + \frac{1}{1.96} \right)$

$1500 = X \left( \frac{1.4 + 1}{1.96} \right)$

$1500 = X \left( \frac{2.4}{1.96} \right)$

Now, isolate $X$ by multiplying both sides by $\frac{1.96}{2.4}$:

$X = 1500 \times \frac{1.96}{2.4}$

To simplify the calculation, we can remove the decimals:

$X = 1500 \times \frac{196}{240}$

Simplify the fraction $\frac{196}{240}$ by dividing the numerator and denominator by their greatest common divisor. Both are divisible by 4:

$\frac{196 \div 4}{240 \div 4} = \frac{49}{60}$

Now, substitute the simplified fraction back into the equation for $X$:

$X = 1500 \times \frac{49}{60}$

Perform the multiplication:

$X = \frac{1500}{60} \times 49$

$X = 25 \times 49$

Calculate the final product:

$25 \times 49 = 25 \times (50 - 1) = 25 \times 50 - 25 \times 1 = 1250 - 25 = 1225$

So, the value of each installment is ₹1225.

Conclusion

The value of each equal annual installment paid by Rajnish to repay the loan of ₹1,500 at a 40% per annum compound interest rate over two years is ₹1225.

Revision Table: Compound Interest Loan Repayment

Let's summarize the key aspects of solving such problems:

Understanding the key terms in a loan repayment problem involving compound interest is crucial.

• Principal ($\text{P}$): The initial amount borrowed.
• Rate ($\text{R}$): The annual interest rate.
• Number of Installments ($\text{n}$): The total number of payments.
• Installment Amount ($X$): The fixed amount paid at regular intervals.
• Compounding Period: The frequency at which interest is added to the principal (usually annually, semi-annually, etc.).

The core principle is equating the present value of all future installments to the original principal.

Additional Information: Alternate Approach (Balance Method)

Another way to solve this is by tracking the outstanding balance:

1. Year 1:
   • Loan amount at start: ₹1500
   • Interest for Year 1: 40% of 1500 = $0.40 \times 1500 = ₹600$
   • Total amount due at end of Year 1 (before payment): $1500 + 600 = ₹2100$
   • First installment paid: ₹$X$
   • Balance carried to Year 2: $2100 - X$
2. Year 2:
   • Loan amount at start of Year 2: $(2100 - X)$
   • Interest for Year 2: 40% of $(2100 - X) = 0.40 \times (2100 - X) = 840 - 0.4X$
   • Total amount due at end of Year 2 (before payment): $(2100 - X) + (840 - 0.4X) = 2100 + 840 - X - 0.4X = 2940 - 1.4X$
   • Second installment paid: ₹$X$
   • The entire balance must be cleared by the second installment: $2940 - 1.4X = X$

Now, solve the equation $2940 - 1.4X = X$ for $X$:

$2940 = X + 1.4X$

$2940 = 2.4X$

$X = \frac{2940}{2.4}$

To perform the division, multiply the numerator and denominator by 10:

$X = \frac{29400}{24}$

Divide 29400 by 24:

$X = 1225$

Both methods yield the same result. The present value method is often more generalizable for a larger number of installments.