The cash price of a refrigerator is Rs. 7044. A customer paid Rs. 2000 in cash and promised to pay the remaining money in 3 yearly equal instalments at the rate of `5%` per annum compound interest. What is the value of each instalment ?

Options

A. Rs.1865
B. Rs.1868.28
C. Rs.1752
D. Rs.1852.20

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the Remaining Amount
The cash price of the refrigerator is Rs. 7044. The customer paid Rs. 2000 in cash.

**Remaining Amount = Cash Price - Cash Paid**
\[ \text{Remaining Amount} = 7044 - 2000 = 5044 \]

### Step 2: Set Up the Equation for Installments
Let the value of each installment be \( P \). The customer will pay this installment in 3 yearly equal installments at a compound interest rate of 5% per annum.

The present value of the installments can be calculated as follows:
1. The first installment \( P \) is paid at the end of the first year.
2. The second installment \( P \) is paid at the end of the second year.
3. The third installment \( P \) is paid at the end of the third year.

The present values of these installments can be calculated using the formula for present value:
\[ PV = \frac{P}{(1 + r)^n} \]
Where \( r \) is the interest rate and \( n \) is the number of years.

Thus, the present value of the installments can be expressed as:
\[ \text{PV of Installments} = \frac{P}{(1 + 0.05)^1} + \frac{P}{(1 + 0.05)^2} + \frac{P}{(1 + 0.05)^3} \]

### Step 3: Substitute Values into the Equation
Substituting the values into the equation:
\[ \frac{P}{1.05} + \frac{P}{(1.05)^2} + \frac{P}{(1.05)^3} = 5044 \]

Calculating the denominators:
- \( (1.05)^1 = 1.05 \)
- \( (1.05)^2 = 1.1025 \)
- \( (1.05)^3 = 1.157625 \)

Thus, the equation becomes:
\[ \frac{P}{1.05} + \frac{P}{1.1025} + \frac{P}{1.157625} = 5044 \]

### Step 4: Factor Out \( P \)
Factoring out \( P \):
\[ P \left( \frac{1}{1.05} + \frac{1}{1.1025} + \frac{1}{1.157625} \right) = 5044 \]

Calculating the individual fractions:
- \( \frac{1}{1.05} \approx 0.95238 \)
- \( \frac{1}{1.1025} \approx 0.90703 \)
- \( \frac{1}{1.157625} \approx 0.86383 \)

Adding these values:
\[ 0.95238 + 0.90703 + 0.86383 \approx 2.72324 \]

### Step 5: Solve for \( P \)
Now we can substitute back into the equation:
\[ P \cdot 2.72324 = 5044 \]
\[ P = \frac{5044}{2.72324} \approx 1852.20 \]

### Final Answer
The value of each installment is approximately **Rs. 1852.20**.

---