RRB Group D 17-Sep-2018 Shift 1 Solved Paper

Let the two numbers be $\mathit{x}$ and $\mathit{y}$. then according to question
$\mathit{x}+\mathit{y}=9$ ........ (i).
also $\frac{1}{\mathit{x}}+\frac{1}{\mathit{y}}=\frac{1}{2}\Rightarrow \mathit{x}+\mathit{y}=\frac{\mathit{x}\mathit{y}}{2}$ ........... (ii)
from (1) and (2)
$\mathit{x}\mathit{y}=18$ ....... (iii)
from (1)
$\mathit{x}=9-\mathit{y}$
putting $\mathit{x}$ in eqn (3)
$\left(9-\mathit{y}\right)\mathit{y}=18$
$\Rightarrow {\mathit{y}}^2-9\mathit{y}+18=0$
$\left(\mathit{y}-6\right)\left(\mathit{y}-3\right)=0$
$\Rightarrow \mathit{y}=6,3$
$\therefore$ One of the no.'s 6 or 3