A loan of Rs 1000 is fully repaid by two installments of Rs 530 and Rs 594, paid at the end of first and second year, respectively. If the interest is compounded annually, then the rate of interest, in percentage, is

Let the annual interest rate be (r) (in decimal). Discount the two instalments to present value:

$$\dfrac{530}{1+r}+\dfrac{594}{(1+r)^2}=1000$$

Set $$x=\dfrac{1}{1+r}$$. Then

$$594x^2+530x-1000=0$$

Discriminant = $$530^2+4\cdot594\cdot1000=280900+2376000=2656900=1630^2$$.

$$x=\dfrac{\ -b\pm\sqrt{b^2-4ac}}{2a}$$

$$x=\dfrac{-530+1630}{2\cdot594}=\frac{1100}{1188}=\dfrac{275}{297}$$

So $$1+r=\dfrac{297}{275}$$

$$r=\dfrac{297-275}{275}=\dfrac{22}{275}=\dfrac{2}{25}=0.08=8\%$$