The percentage scrap in a sheet metal blanking operation of a continuous strip of sheet metal as shown in the figure is _______ (correct to two decimal places). 

The correct answer is 53.25.

Based on the provided figure of the continuous sheet metal strip blanking operation, we can identify the following parameters:
1. Diameter of each blanked circular disc = $D$
2. Margin on both the top and bottom sides of the strip = $\frac{D}{5}$
3. Bridge thickness (distance between two consecutive circular blanks) = $\frac{D}{5}$

Let us determine the total width ($W$) of the sheet metal strip:
$W=D+\frac{D}{5}+\frac{D}{5}=\frac{7D}{5}$

The feed or pitch ($P$) which is the distance between the centers of two consecutive blanked holes is given by:
$P=D+\frac{D}{5}=\frac{6D}{5}$

For each repeating rectangular unit of the strip of length equal to the pitch ($P$) and width equal to the strip width ($W$), one full circular blank is produced.
Thus, the total area of this repeating unit ($A_t$) is:
$A_t=W\times P=\frac{7D}{5}\times \frac{6D}{5}=\frac{42D^2}{25}$

The useful area (area of the blanked disc, $A_u$) is:
$A_u=\frac{\pi D^2}{4}$

The percentage of scrap is calculated as the ratio of scrap area to the total area:
$\text{\% of scrap}=\frac{A_t-A_u}{A_t}\times 100\text{\%}=\left(1-\frac{A_u}{A_t}\right)\times 100\text{\%}$

Substituting the values of $A_u$ and $A_t$:
$\text{\% of scrap}=\left(1-\frac{\frac{\pi }{4}}{\frac{42}{25}}\right)\times 100\text{\%}$
$\text{\% of scrap}=\left(1-\frac{25\pi }{168}\right)\times 100\text{\%}$
$\text{\% of scrap}\approx \left(1-0.4675\right)\times 100\text{\%}=53.25\text{\%}$