Question

A number consists of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.

Answer 1

Let the digits in the tens be $x$ and ones place be $y$.

Hence the number is $10x+y$

By reversing $10y+x$

Sum of digit $=5$

$\Rightarrow x+y=5⟶\left(i\right)$

Also that when $9$ is added to the number the digits get interchanged.

$\therefore (10x+y)+9=(10y+x)$

$10x+y+9=10y+x=0$

$9x-9y=-9$

$x-y=-1⟶\left(ii\right)$

Adding $\left(i\right)$ &$\left(ii\right)$ we get ,

$\Rightarrow x+y=5$

$\Rightarrow x-y=-1$

$\Rightarrow 2x=4$

$\Rightarrow x=2$

Put $x=2$ in $x+y=5$

$\therefore 2+y=5$

$\Rightarrow y=3$

Hence the no: is $23$.