A sum of money becomes $$\frac{7}{6}$$ of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is ?

$$\eqalign{ & {\text{Let principal = 6P}} \cr & {\text{Hence Amount}} \cr & {\text{ = 6P}} \times \frac{7}{6} = 7{\text{P}} \cr & \therefore {\text{SI = 7P}} - {\text{6P = P}} \cr & {\text{Time = 3 years}} \cr & \boxed{{\text{SI = }}\frac{{{\text{P}} \times {\text{T}} \times {\text{R}}}}{{100}}} \cr & \Rightarrow {\text{P = }}\frac{{6{\text{P}} \times {\text{R}} \times {\text{3}}}}{{100}} \cr & \Rightarrow {\text{R = }}\frac{{100}}{{18}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50}}{9} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 5\frac{5}{9}\% \cr} $$

Alternate:
Note: In such type of questions to save your valuable time try to think like that.
\[\begin{gathered} \begin{array}{*{20}{c}} {{\text{Amount}}}&{{\text{Principal}}} \end{array} \hfill \\ {\text{ }}\underbrace {7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{6}}}_{ + 1} \hfill \\ \end{gathered} \]
$$\eqalign{ & {\text{Required rate % }} \cr & {\text{ = }}\frac{1}{6} \times \frac{{100}}{3} \cr & = 5\frac{5}{9}\% \cr} $$