Let the amount invested by Anil be ₹$\text{P}$, at the rate of interest $r$ annually.
The amount at the end of:
- First year $= \text{P}(1+r)$
- Second year $= \text{P}(1+r)^{2}$
- Third year $= \text{P}(1+r)^{3}$
- Fourth year $= \text{P}(1+r)^{4}$
Now, $\text{P}(1+r)^{2} – \text{P}(1+r) = 806.25$
$\Rightarrow \text{P}(1+r) [(1+r)-1] = 806.25$
$\Rightarrow \text{P}(1+r)\cdot r = 806.25\quad \longrightarrow(1)$
And, $\text{P}(1+r)^{3} – \text{P}(1+r)^{2} = 866.72$
$\Rightarrow \text{P}(1+r)^{2}[(1+r)-1] = 866.72$
$\Rightarrow \text{P}(1+r)^{2}\cdot r = 866.72\quad \longrightarrow(2)$
Divide the equation $(2)$ by equation $(1).$
$\Rightarrow \dfrac{\text{P}(1+r)^{2}\cdot r}{\text{P}(1+r)\cdot r} = \dfrac{866.72}{806.25}$
$\Rightarrow (1+r) = \dfrac{866.72}{806.25}\quad \longrightarrow(3)$
The interest accrued, during the fourth year $= \text{P}(1+r)^{4} – \text{P}(1+r)^{3}$
$\qquad = \text{P}(1+r)^{3}[(1+r)-1]$
$\qquad = \text{P}(1+r)^{3}\cdot r$
$\qquad = \text{P}(1+r)^{2}\cdot r \cdot (1+r)$
$\qquad = 866.72 \times\dfrac{866.72}{806.25}\quad [\because \text{from equation (2) and (3)}]$
$\qquad =$ ₹ $931.72$
Correct Answer $:\text{D}$