The value of

​$$\tfrac{4}{6}+\tfrac{1}{6}+\tfrac{4}{12} +\tfrac{3}{6}-1=$$​​

Correct option is C

Given:

​$$\tfrac{4}{6}+\tfrac{1}{6}+\tfrac{4}{12}+\tfrac{3}{6}-1=$$​​

Formula Used:

​$$\begin {array}{|c|c|} \hline \textbf{Operation preference wise} & \textbf{Symbol} \\ \hline Brackets &[],{}, () \\ \hline Orders, of & ² (power), √ (root) , of \\ \hline Division & ÷ \\ \hline Multiplication & × \\ \hline Addition & + \\ \hline Subtraction & - \\ \hline \end{array}$$​

Solution:

​$$\tfrac{4}{6}+\tfrac{1}{6}+\tfrac{4}{12}+\tfrac{3}{6}-1$$​​

= $$\tfrac{8+2+4+6}{12} - 1$$​​

​$$= \tfrac{20}{12} -1$$​​

​$$= \tfrac{20-12}{12}$$​​

​$$=\tfrac{8}{12}$$​​