SSC CGL 31 Aug 2016 Paper

${\mathit{l}}^2+{\mathit{m}}^2+{\mathit{n}}^2$ = 31;
l + m + n = 9
On squaring both sides,
${\left(\mathit{l}+\mathit{m}+\mathit{n}\right)}^2$ = 81
⇒ ${\mathit{l}}^2+{\mathit{m}}^2+{\mathit{n}}^2$ + 2 (lm + mn + nl) = 81
⇒ 31 + 2(lm + mn + nl) = 81
⇒ 2(lm + mn + nl) = 81 – 31 = 50
⇒ lm + mn + nl = $\frac{50}{2}$ = 25