Question 10.

Let the principal amount be $P$ and the interest rate be $r$.

Then

$P(1+r)^2 - P(1+r) = 806.25 \quad (1)$

$P(1+r)^3 - P(1+r)^2 = 866.72 \quad (2)$

Dividing (2) by (1):

$\dfrac{P(1+r)^3 - P(1+r)^2}{P(1+r)^2 - P(1+r)} = \dfrac{866.72}{806.25}$

$\dfrac{(1+r)^2 - 1 - r}{1+r - 1} = 1.075$

$\dfrac{r^2 + r}{r} = 1.075$

$r = 0.075$ or $7.5$

$\frac{\text{Interest accrued in 4th year}}{\text{Interest accrued in 3rd year}}$ = $\dfrac{X}{866.72}$

$\dfrac{P(1+r)^4 - P(1+r)^3}{P(1+r)^3 - P(1+r)^2} = \dfrac{X}{866.72}$

Dividing numerator and denominator by $P(1+r)^2$:

$\dfrac{r^2 + 2r + 1 - 1 - r}{r} = \dfrac{X}{866.72}$

$r + 1 = \dfrac{X}{866.72}$

$X = 1.075 \times 866.72 = 931.72$