Question

The sum of the digits of a two-digit number is $12$. If the digits are reversed, the new number is $12$ less than twice the original number. Find the original number.

Answer 1

Let the two digit number be $10x+y$

Sum of digit : $x+y=12$ __(i)

A.t.Q : $2(10x+y)-(10y+x)=12$

$\Rightarrow 20x+2y-10y-x=12$

$\Rightarrow 19x-8y=12$

$\Rightarrow 19x-8y-12=0$ __(ii)

Put eq. (i) in eq. (ii)

$\Rightarrow 19(12-y)-8y-12=0$

$\Rightarrow 228-19y-8y-12=0$

$\Rightarrow 27y=216$

$\Rightarrow y=8$

$x+y=12$

$x=12-8$

$x=4$

Number : $48$