$\begin{matrix} 13 & (27) & 55 11 & (23) & 47 7 & (. . . .) & 31 \end{matrix}$

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Solution

The correct option is B 15
The number in the second column is one more than twice the number in first column. Similarly, the number in third column is one more than twice the number in second column.
As given, $2 \times 13 + 1 = 27$
$2 \times 27 + 1 = 55$
$2 \times 11 + 1 = 23$
$2 \times 23 + 1 = 47$
Thus, the missing number is: $2 \times 7 + 1 = 15$

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