Let Anil invest a principal amount \( P \) at an interest rate \( r\% \) per annum, compounded annually. The interest accrued during the second and third years are given, and we are tasked with finding the interest accrued during the fourth year.
The interest accrued during the second year is: \[ P \left( 1 + \frac{r}{100} \right)^2 - P \] This is equal to 806.25 rupees. Thus, we have the equation: \[ P \left( 1 + \frac{r}{100} \right)^2 = P + 806.25 \quad \text{(1)} \]
The interest accrued during the third year is: \[ P \left( 1 + \frac{r}{100} \right)^3 - P \left( 1 + \frac{r}{100} \right)^2 \] This is equal to 866.72 rupees, so we have the equation: \[ P \left( 1 + \frac{r}{100} \right)^3 = P \left( 1 + \frac{r}{100} \right)^2 + 866.72 \quad \text{(2)} \] Substituting the value from equation (1) into equation (2): \[ P \left( 1 + \frac{r}{100} \right)^3 = P + 806.25 + 866.72 \] Simplifying: \[ P \left( 1 + \frac{r}{100} \right)^3 = P + 1672.97 \]
The ratio of the interest accrued in the third year to the interest accrued in the second year is given by: \[ \frac{866.72}{806.25} = 1 + \frac{r}{100} \] Solving this equation: \[ 1 + \frac{r}{100} = 1.075 \] \[ \frac{r}{100} = 0.075 \] \[ r = 7.5\% \]
Now that we know \( r = 7.5\% \), we can substitute this value into the formula for the fourth year: \[ \text{Interest during the fourth year} = P \left( 1 + \frac{7.5}{100} \right)^4 - (P + 1672.97) \] Simplifying: \[ \text{Interest during the fourth year} = P \left( 1.075^4 \right) - (P + 1672.97) \] We also know from the second year that: \[ 806.25 = P \left( 1.075^2 - 1 \right) \] Using this, we can find the value of \( P \) and subsequently compute the interest for the fourth year.
The calculation can get complex and may require either iterative approaches or some algebraic simplifications. However, based on the increasing trend of interest in each subsequent year, the interest for the fourth year will be higher than the third year, which is higher than the second year. Among the given options, the interest during the fourth year is closest to 931.72 rupees.
The correct option is \( \boxed{(A): 931.72} \).
Let \( P \) be the principal amount and \( r \) be the interest rate (in decimal form).
Divide equation (ii) by (i):
\[ \frac{P(1+r)^3 - P(1+r)^2}{P(1+r)^2 - P(1+r)} = \frac{866.72}{806.25} \] Simplify numerator and denominator by factoring out \( P(1+r) \): \[ \frac{P(1+r)^2 \left[(1+r) - 1\right]}{P(1+r)\left[(1+r) - 1\right]} = \frac{866.72}{806.25} \] Which reduces to: \[ \frac{(1+r)^2 - (1+r)}{(1+r) - 1} = 1.075 \] Expand numerator: \[ \frac{(1 + 2r + r^2) - (1 + r)}{r} = 1.075 \] \[ \frac{r^2 + r}{r} = 1.075 \] \[ r + 1 = 1.075 \implies r = 0.075 = 7.5\% \] ---
Interest in 4th year, \( X \), satisfies:
\[ \frac{P(1+r)^4 - P(1+r)^3}{P(1+r)^3 - P(1+r)^2} = \frac{X}{866.72} \] Divide numerator and denominator by \( P(1+r)^2 \): \[ \frac{(1+r)^2 - (1+r)}{(1+r) - 1} = \frac{X}{866.72} \] From previous step, numerator simplifies to: \[ r + 1 = 1.075 \] Thus: \[ \frac{X}{866.72} = 1.075 \implies X = 1.075 \times 866.72 = 931.72 \] ---
The interest for the 4th year is \(\boxed{931.72}\).
Therefore, the correct option is (A): 931.72