Question

Suppose $x$ and $y$ are positive real numbers such that $x\sqrt{x}+y\sqrt{y}=183$ and $x\sqrt{y}+y\sqrt{x}=182$ then value of $\frac{18}{5}(x+y)$ is :

• A. $73$
• B. $146$
• C. $63$
• D. $126$

Answer 1

The correct option is B $146$

$x\sqrt{y}+y\sqrt{x}=182$
$\sqrt{x}\sqrt{y}(\sqrt{x}+\sqrt{y})=182....\left(1\right)$

$x\sqrt{x}+y\sqrt{y}=183$
$(\sqrt{x}{)}^3+(\sqrt{y}{)}^3=183$

$(\sqrt{x}+\sqrt{y}{)}^3-3\sqrt{xy}(\sqrt{x}+\sqrt{y})=183$

$\Rightarrow (\sqrt{x}+\sqrt{y}{)}^3-3\times 182=183.....................[From\left(1\right)]$

$\Rightarrow \sqrt{x}+\sqrt{y}=9$

$\therefore \frac{18}{5}(x+y)=\frac{18}{5}\left(\right(\sqrt{x}+\sqrt{y}{)}^2-2\sqrt{xy})$

$=\frac{18}{5}\left(\frac{365}{9}\right)$

$=146$.