Given that,
- $x^{2} + y^{2} = 0.1 \quad \longrightarrow (1)$
- $\mid x-y \mid = 0.2 \quad \longrightarrow (2) $
We know that, ${\color{Green}{\mid x \mid = \left\{\begin{matrix} x \:;&x\geq 0 \\ -x\:; &x<0 \end{matrix}\right.}}$
Now, from equation $(2),$ we get
We know that, ${\color{Blue}{(x-y)^{2} = x^{2} + y^{2} - 2xy}}$
$\Rightarrow (0.2)^{2} = (-0.2)^{2} = 0.1 - 2xy$
$\Rightarrow 0.04 = 0.1 -2xy$
$\Rightarrow 2xy = 0.1 -0.04 = 0.06$
$\Rightarrow xy = 0.03$
$\therefore \boxed{ \mid xy \mid = \mid 0.03 \mid = 0.03}$
Now, $(\mid x \mid + \mid y \mid)^{2} = \mid x \mid ^{2} + \mid y \mid ^{2} + 2\mid x \mid \mid y \mid $
$\Rightarrow (\mid x \mid + \mid y \mid)^{2} = \underbrace{x^{2} + y^{2}}_{0.1} + 2 \; \underbrace{\mid xy \mid}_{0.03}$
$\Rightarrow (\mid x \mid + \mid y \mid)^{2} = 0.1 + 2(0.03) = 0.1 + 0.06 = 0.16$
$\Rightarrow \mid x \mid + \mid y \mid = \sqrt{0.16} $
$\Rightarrow \boxed{ \mid x \mid + \mid y \mid = 0.4}$
Correct Answer $:\text{B}$