Delhi Police Constable 2014 Paper

Let greater no. = x and smaller no. = y
According to the question,
$\mathit{x}\times \mathit{y}=8091$
$\mathit{x}-\mathit{y}=6$ → (1)
$\therefore {\left(\mathit{x}+\mathit{y}\right)}^2={\left(\mathit{x}-\mathit{y}\right)}^2+4\mathit{x}\mathit{y}$
$={\left(6\right)}^2+4\times 8091$
$=36+32364$
$=32400$
$\therefore \mathit{x}+\mathit{y}=\sqrt{32400}$
$=180$ → (2)
From equation (1) and (2)
$\mathit{x}+\mathit{y}=180$
$\mathit{x}-\mathit{y}=6$
$2\mathit{x}=186$
$\therefore \mathit{x}=\frac{186}{2}$
$=93$
$\mathit{y}=180-93$
$=87$
Hence, smaller no will be 87.