SSC CHSL 8 Jan 2017 Shift 1 Paper

Let the unit's digit of the number be and ten's digit be x
⇒ Number = $10\mathit{x}+\mathit{y}$
Product of digits = $\mathit{x}\mathit{y}=18$ ---(i)
According to question, ⇒ $10\mathit{x}+\mathit{y}+63=10\mathit{y}+\mathit{x}$
⇒ $9\mathit{y}-9\mathit{x}=63$
⇒ $\mathit{y}-\mathit{x}=\frac{63}{9}=7$ ------(ii)
Substituting value of y from equation (ii) in (i), we get :
⇒ $\mathit{x}\left(7+\mathit{x}\right)=18$
⇒ ${\mathit{x}}^2+7\mathit{x}-18=0$
⇒ ${\mathit{x}}^2+9\mathit{x}-2\mathit{x}-18=0$
⇒ $\mathit{x}\left(\mathit{x}+9\right)-2\left(\mathit{x}+9\right)=0$
⇒ $\mathit{x}=2,-9$
Since x is a digit and can't be negative, ⇒ $\mathit{x}=2$
Substituting it in equation (ii), ⇒ $\mathit{y}=7+2=9$
∴ Number = 29