34 views If $x^{2}+y^{2}= 0.1$ and $\left | x-y \right |=0.2$, then $\left | x \right |+\left | y \right |$ is equal to

1. $0.3$
2. $0.4$
3. $0.2$
4. $0.6$

recategorized | 34 views

Given that $:x^{2} + y^{2} = 0.1\rightarrow(1)$

and $\mid x-y \mid = 0.2\rightarrow(2)$

We know that $\mid x \mid = \left\{\begin{matrix} x \:;&x\geq 0 \\ -x\:; &x<0 \end{matrix}\right.$

Now, $x -y = 0.2$ and $x-y = -0.2 \rightarrow(3)$

We know that $:(x-y)^{2} = x^{2} + y^{2} - 2xy$

$\implies (0.2)^{2} = (-0.2)^{2} = 0.1 - 2xy$

$\implies 0.04 = 0.1 -2xy$

$\implies 2xy = 0.1 -0.04 = 0.06$

$\implies xy = 0.03$

$\therefore\: \mid xy \mid = \mid 0.03 \mid = 0.03$

Now, $(\mid x \mid + \mid y \mid)^{2} = \mid x \mid ^{2} + \mid y \mid ^{2} + 2\mid x \mid \mid y \mid$
$\implies (\mid x \mid + \mid y \mid)^{2} = x^{2} + y^{2} + 2 \mid xy \mid$

$\implies (\mid x \mid + \mid y \mid)^{2} = 0.1 + 2(0.03) = 0.1 + 0.06 = 0.16$

$\implies \mid x \mid + \mid y \mid = 0.4$

So, the correct answer is $(B).$
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