CAT 2015 | Question: 79

Question

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

• A. $0.3$
• B. $0.4$
• C. $0.2$
• D. $0.6$

Answer 1

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

• $x -y = 0.2$
• $x-y = -0.2$

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 $:\text{B}$