If ‘A’ stands for ‘÷’, ‘B’ stands for ‘×’, ‘C’ stands for ‘+’ and ‘D’ stands for ‘−’, then the resultant of which of the following would be 77?​

A → ÷ , B → × , C → +, D → −

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

We must find which option gives a result = 77

Option A:

42 B 6 C 38 D 18 A 6

= 42 × 6 + 38 − 18 ÷ 6

Apply BODMAS rule:

42 × 6 = 252

18 ÷ 6 = 3

Now expression = 252 + 38 − 3

= 290 − 3

= 287 (Incorrect)

Option B:

42 A 6 D 38 C 18 B 6

= 42 ÷ 6 − 38 + 18 × 6

42 ÷ 6 = 7

18 × 6 = 108

Now = 7 − 38 + 108

= (7 − 38) + 108

= −31 + 108

Option C:

42 C 6 B 38 A 18 D 6

= 42 + 6 × 38 ÷ 18 − 6

6 × 38 = 228

228 ÷ 18 = 12.666...

Now = 42 + 12.666 − 6 = 48.666 (Incorrect)

Option D:

42 D 6 A 38 B 18 C 6

= 42 − 6 ÷ 38 × 18 + 6

6 ÷ 38 = 0.1579

0.1579 × 18 = 2.842

Now = 42 − 2.842 + 6 = 45.158 (Incorrect)

Thus, correct option is (b).