Question:
Sum of the digits of a two-digit number is 11. The given number is less than the number obtained by interchanging the digits by 9. Find the number.
Solution:
Let the unit's digit be $x$. Then, the ten's digit $=11-x$
$\therefore$ Number $=10(11-x)+x=110-10 x+x=110-9 x$
Number obtained by interchanging the digits $=10 x+(11-x)=9 x+11$
According to the question,
$9 x+11-(110-9 x)=9$
$\Rightarrow$ $9 x+11-110+9 x=9$
$\Rightarrow \quad 18 x=9-11+110$
$\Rightarrow$ $18 x=108$
$\Rightarrow$ $x=\frac{108}{18}$
$\Rightarrow$ $x=6$
Hence, unit's digit $=6$ and ten's digit $=11-6=5$ Therefore, the required number is 56 .