Sainik School Entrance Exam 2018 Solved Paper

(d) Suppose ten's place digit number be $\mathit{x}$ and unit's place digit number be $\mathit{y}$
∴ Number = $10\mathit{x}+\mathit{y}$
⇒ $\mathit{x}=\mathit{y}+4$
⇒ $\mathit{x}-\mathit{y}=4$
⇒ $\mathit{x}+\mathit{y}=\frac{1}{7}\left(10\mathit{x}+\mathit{y}\right)$
⇒ $7\mathit{x}+7\mathit{y}=10\mathit{x}+\mathit{y}$
⇒ $3\mathit{x}-6\mathit{y}=0$
From (1) and (2)
$\mathit{x}-\mathit{y}=4]\times 3$
$3\mathit{x}-6\mathit{y}=0]\times 1$
$3\mathit{x}-3\mathit{y}=12$
$3\mathit{x}-6\mathit{y}=0$
$\frac{-+-}{3\mathit{y}=12}$
⇒ $\mathit{y}=4$
∴ $\mathit{x}-\mathit{y}=4$
$\mathit{x}-4=4$
⇒ $\mathit{x}=8$
Hence, number $=10\mathit{x}+\mathit{y}=84$.