Correct option is CGiven: 1035÷[34(71+65)−1534]1035 \div \left[ \frac{3}{4}(71+65)-15\frac{3}{4}\right]1035÷[43(71+65)−1543] Concept Used: BODMAS rule; Operation preference wiseSymbolBrackets[],,()Orders,of²(power),√(root),ofDivision÷Multiplication×Addition+Subtraction−\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} Operation preference wiseBracketsOrders,ofDivisionMultiplicationAdditionSubtractionSymbol[],,()²(power),√(root),of÷×+− Solution: =1035÷[34(71+65)−1534] =1035÷[34(136)−634] =1035÷[4084−634] =1035÷[3454] =1035×4345 =3×4 =12 = 1035 \div \left[ \frac{3}{4}(71+65)-15\frac{3}{4}\right] \\ \ \\ = 1035 \div \left[ \frac{3}{4}(136)-\frac{63}{4}\right] \\ \ \\ = 1035 \div \left[ \frac{408}{4}-\frac{63}{4}\right] \\ \ \\ = 1035 \div \left[ \frac{345}{4}\right] \\ \ \\ = 1035 \times \frac{4}{345} \\ \ \\ = 3 \times 4 \\ \ \\ = \bf 12 =1035÷[43(71+65)−1543] =1035÷[43(136)−463] =1035÷[4408−463] =1035÷[4345] =1035×3454 =3×4 =12