Simplify:

​$$0.3\div\lparen0.6×\frac{2}{3}\rparen×0.2\lparen0.2×2\div\frac{1}{2}+0.2\rparen$$​​

Correct option is C

Given:

 $$0.3\div\lparen0.6×\frac{2}{3}\rparen×0.2\lparen0.2×2\div\frac{1}{2}+0.2\rparen$$

Concept Used: 

BODMAS rule: 

​$$\begin{array}{|c|c|} \hline \textbf{Operation preference wise} & \textbf{Symbol} \\ \hline Brackets &[],{}, () \\ \hline Orders, of & x(power), √ (root) , of \\ \hline Division & ÷ \\ \hline Multiplication & × \\ \hline Addition & + \\ \hline Subtraction & - \\ \hline \end{array}$$​​

Solution:

$$0.3\div\lparen0.6×\frac{2}{3}\rparen×0.2\lparen0.2×2\div\frac{1}{2}+0.2\rparen$$

= $$0.3\div\lparen {0.4}\rparen×0.2\times\lparen0.2×2\times 2+0.2\rparen$$

= $$\frac{3}{4}×0.2\times\lparen0.8+0.2\rparen$$

= $$\frac{3}{40}×2\times1$$

= $$\frac{3}{20}$$

= 0.15

Thus, the correct answer is (c)