Given: Regular polygons A and B possess interior angles in the ratio \(3 : 4\).
Let polygon A have \(n\) sides. According to the problem statement, polygon B has twice the number of sides as polygon A, meaning it has \(2n\) sides.
The formula for the interior angle of a regular polygon with \(k\) sides is:
\[\text{Interior angle} = \frac{(k - 2) \times 180^\circ}{k}\]For polygon A, the interior angle is \(\frac{(n - 2) \times 180}{n}\).
For polygon B, the interior angle is \(\frac{(2n - 2) \times 180}{2n}\).
The given ratio of their interior angles is:
\[\frac{(n - 2) \times 180}{n} : \frac{(2n - 2) \times 180}{2n} = 3 : 4\]Divide both sides of the ratio by 180:
\[\frac{n - 2}{n} : \frac{2n - 2}{2n} = 3 : 4\]Applying cross-multiplication to the ratio:
\[4 \times \frac{n - 2}{n} = 3 \times \frac{2n - 2}{2n}\]This simplifies to:
\[\frac{4(n - 2)}{n} = \frac{3(2n - 2)}{2n}\]Multiply both sides by \(2n\) to eliminate denominators:
\[2n \times \frac{4(n - 2)}{n} = 2n \times \frac{3(2n - 2)}{2n}\] \[8(n - 2) = 3(2n - 2)\]Expand both sides:
\[8n - 16 = 6n - 6\]Rearrange terms to solve for \(n\):
\[8n - 6n = -6 + 16\] \[2n = 10\] \[n = 5\]The initial attempts at solving contained algebraic errors leading to invalid results for \(n\).
With \(n = 5\), polygon A has 5 sides.
The number of sides of polygon B is \(2n\), which is \(2 \times 5 = 10\).
Therefore, Polygon A has 5 sides, and Polygon B has \(\boxed{10}\) sides.