Question 56

# 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 INR, during the fourth year is nearest to

Solution

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

Then $$P\left(1+r\right)^2-P\left(1+r\right)=806.25$$ -(1)

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

Dividing (2) by (1), we get:

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

$$\frac{\left(\left(1+r\right)^2-1-r\right)}{1+r-1}=1.075$$

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

r=0.075 or 7.5%

$$\frac{\left(Interest\ accrued\ in\ 4th\ yr\right)}{Interest\ accrued\ in\ 3rd\ yr}=\frac{X}{866.72}$$

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

Dividing numerator and denominator by $$P\left(1+r\right)^2$$

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

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

$$X=1.075\times\ 866.72=931.72$$

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