Question

The digits of two - digits number is differ by $5$ . If the digits are interchanged and the resulting number is added to the original number , we get $99$ . Find the original number .

Answer 1

Let the digit in ten's$.$ Place be $x$ and the digit in the one's place be $y.$

$\therefore x-y=5⟶\left(i\right)$

$\therefore$ Two digit number$=10x+y$

Two digit after reversing the digits$=10y+x$

According to question $,$

$\therefore 10x+y+10y+x=99$

$\Rightarrow 11x+11y=99$

$\Rightarrow x+y=9⟶\left(ii\right)$

On adding $\left(i\right)$ and $\left(ii\right),$

$2x=14$

$\Rightarrow x=7$

Putting the value of $x$ in equation $\left(i\right)$

$x-y=5$

$\Rightarrow y=2$

$\therefore$ Number$=10x+y$

$=72.$