Question

The sum of digits of a two-digit number is 7. If the digits are reversed, the new number decreased by 2, equals twice the original number. Find the number.

Answer 1

Let the digit at unit's place be x and the digit ten's place y.
Required no $=10y+x$
If the digit's are reversed,
reversed no = $10y+x$
According to the question,
$x+y=\mathrm{7...}\left(1\right)$
and
$10x+y-2=2(10y+x)$
$8x-19y=2..\left(2\right)$
multiplying equation no (1) by $19$
Now adding the two equation,
$19x+19y=\mathrm{133........}\left(3\right)$
$8x-19y=\mathrm{2............}\left(2\right)$
__________________________
$27x=135$
$x=5$
From (1)
$5+y=7$
$y=2$
Required no is
$10\left(2\right)+5=25$