What annual installment will discharge a debt of Rs. 9,600 due in 5 years at 10% simple interest?
Rs. 1600
This problem asks us to find the equal annual installment amount needed to pay off a debt over a specific period, considering simple interest is applied to the outstanding balance or calculated based on the total principal and time.
In a simple interest installment plan, when you pay an installment, it reduces the amount you owe. However, the interest for each period is typically calculated on the outstanding amount at the beginning of that period. Alternatively, and more commonly for simple interest installments as in this problem, we can think of the total debt as the sum of the values of each installment accumulated with simple interest up to the final payment date.
Let the equal annual installment be \( x \). These installments are paid at the end of each year for 5 years. We can calculate the value of each installment at the end of the 5-year period, including the simple interest it would have earned from the time of payment until the end of the 5th year. The sum of these accumulated values must equal the total debt amount due.
The interest is 10% simple interest per annum. An installment paid at the end of year 'k' will earn simple interest for \((5 - k)\) years.
| Installment No. | Year of Payment End | Time Period for Interest (Years) | Value at End of 5 Years |
|---|---|---|---|
| 1st | Year 1 | \(5 - 1 = 4\) | \(x + \text{SI on } x \text{ for 4 years}\) |
| 2nd | Year 2 | \(5 - 2 = 3\) | \(x + \text{SI on } x \text{ for 3 years}\) |
| 3rd | Year 3 | \(5 - 3 = 2\) | \(x + \text{SI on } x \text{ for 2 years}\) |
| 4th | Year 4 | \(5 - 4 = 1\) | \(x + \text{SI on } x \text{ for 1 year}\) |
| 5th | Year 5 | \(5 - 5 = 0\) | \(x + \text{SI on } x \text{ for 0 years}\) |
Simple Interest (SI) on an amount P at R% per annum for T years is given by \( \text{SI} = \frac{P \cdot R \cdot T}{100} \). Here, P = \( x \) and R = 10.
The sum of these values must equal the total debt due, which is Rs. 9,600.
\( 1.4x + 1.3x + 1.2x + 1.1x + x = 9600 \)
\( (1.4 + 1.3 + 1.2 + 1.1 + 1)x = 9600 \)
\( 6x = 9600 \)
To find the value of \( x \), we divide the total debt by the sum of the multipliers:
\( x = \frac{9600}{6} \)
\( x = 1600 \)
Therefore, the annual installment will be Rs. 1,600.
We can quickly verify this using the formula: Total Amount Due = \( nx + \frac{xr}{100} \frac{n(n-1)}{2} \)
Here, A = 9600, n = 5, r = 10, x = 1600.
\( 9600 = 5 \cdot 1600 + \frac{1600 \cdot 10}{100} \frac{5(5-1)}{2} \)
\( 9600 = 8000 + 160 \cdot \frac{5 \cdot 4}{2} \)
\( 9600 = 8000 + 160 \cdot \frac{20}{2} \)
\( 9600 = 8000 + 160 \cdot 10 \)
\( 9600 = 8000 + 1600 \)
\( 9600 = 9600 \)
The calculation is correct.
The annual installment required to discharge a debt of Rs. 9,600 due in 5 years at 10% simple interest is Rs. 1,600.
| Term | Definition | Formula (for Principal P, Rate R%, Time T years) |
|---|---|---|
| Principal (P) | The initial amount borrowed or lent. | - |
| Rate (R) | The percentage at which interest is charged per period (usually per year). | - |
| Time (T) | The duration for which the money is borrowed or lent. | - |
| Simple Interest (SI) | Interest calculated only on the principal amount. | \( \text{SI} = \frac{P \cdot R \cdot T}{100} \) |
| Amount (A) | The total sum including principal and simple interest. | \( A = P + \text{SI} = P \left(1 + \frac{R \cdot T}{100}\right) \) |
Installments are periodic payments made to repay a debt or loan. There are different ways installment plans are structured, especially concerning how interest is calculated.
Understanding whether a problem involves simple or compound interest is crucial for choosing the correct approach and formula for calculating installments or the total amount due.
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