SSC CHSL 12 Apr 2021 Shift 2 Paper

Let the two numbers be $\mathit{x}$ and $\mathit{y}$ and $\mathit{x}>\mathit{y}$.
$\therefore$ According to the question,
$\mathit{x}-\mathit{y}=5$ . . . (i)
. . . (ii)
$\therefore {\mathit{x}}^3-{\mathit{y}}^3={\left(\mathit{x}-\mathit{y}\right)}^3+3\mathit{x}\mathit{y}\left(\mathit{x}-\mathit{y}\right)$
$1850={\left(5\right)}^3+3\mathit{x}\mathit{y}\times 5$

$\mathit{x}\mathit{y}=115$
Now, we know, ${\left(\mathit{x}-\mathit{y}\right)}^2={\mathit{x}}^2+{\mathit{y}}^2-2\mathit{x}\mathit{y}$
$\therefore {\left(5\right)}^2={\mathit{x}}^2+{\mathit{y}}^2-2\times 115$
$\Rightarrow {\mathit{x}}^2+{\mathit{y}}^2=255$
$\therefore {\left(\mathit{x}+\mathit{y}\right)}^2={\mathit{x}}^2+{\mathit{y}}^2+2\mathit{x}\mathit{y}$
$=255+2\times 115$
$\mathit{x}+\mathit{y}=\sqrt{485}$ . . . (iii)
From Eqs. (i) and (iii);
$\left(\mathit{x}-\mathit{y}\right)\left(\mathit{x}+\mathit{y}\right)=5\times \sqrt{485}$
$\left({\mathit{x}}^2-{\mathit{y}}^2\right)=5\sqrt{485}$