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Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are $₹ \; 806.25$ and $₹ \; 866.72,$ respectively, the interest accrued, in $\text{INR},$ during the fourth year is nearest to

1. $934.65$
2. $929.48$
3. $926.84$
4. $931.72$

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 $(1).$

$\Rightarrow \frac{\text{P}(1+r)^{2}\cdot r}{\text{P}(1+r)\cdot r} = \frac{866.72}{806.25}$

$\Rightarrow (1+r) = \frac{866.72}{806.25}\quad \longrightarrow(3)$

The interest accrued, during the fourth year $= \text{P}(1+r)^{4} – \text{P}(1+r)^{3}$

$\qquad = \text{P}(1+r)^{3}[(1+r)-1]$

$\qquad = \text{P}(1+r)^{3}\cdot r$

$\qquad = \text{P}(1+r)^{2}\cdot r \cdot (1+r)$

$\qquad = 866.72 \times\frac{866.72}{806.25}\quad [\because \text{from equation (2) and (3)}]$

$\qquad = ₹ 931.72$

Correct Answer $:\text{D}$

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