Find the rates of interest per year, if the interest charged for 8 months be 0.06 times of the money borrowed.

Given:

T = 8 months

= $\dfrac{8}{12}$ years

= $\dfrac{2}{3}$ years

Let principal be P

S.I. = 0.06 P

Let rate of interest be $r$.

$\because \text{S.I.} = ₹ \Big(\dfrac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{0.06P} = ₹ \Big(\dfrac{P \times r \times 2}{3 \times 100}\Big)\\[1em] \Rightarrow \dfrac{6P}{100} = ₹ \Big(\dfrac{P \times r \times 2}{300}\Big)\\[1em] \Rightarrow \dfrac{6}{100}\cancel {P} = ₹ \Big(\dfrac{\cancel {P} \times r \times 2}{300}\Big)\\[1em] \Rightarrow \dfrac{6}{100} = ₹ \dfrac{2r}{300}\\[1em] \Rightarrow r = \dfrac{6 \times 300}{100 \times 2}\\[1em] \Rightarrow r = \dfrac{1800}{200}\\[1em] \Rightarrow r = 9 \%$

Hence, the rate of interest = 9%.