Question

A number consist of two digits whose sum is $11$. The number formed by reversing the digits is $9$ less than the original number. Find the number.

Answer 1

Let the tens digit be and the units digit be $y$. Then the number is $10x+y$.
Sum of the digits is $x+y=11$.
The number formed by reversing the digits is $10y+x$.
Given data, $(10x+y)-9=10y+x$
$\Rightarrow 10x+y-10y-x=9$
$9x-9y=9$
Dividing by 9 on both sides, $x-y=1$ ........ (2)
Equation (2) becomes $x=1+y$ .......... (3)
Substituting x in (1) we get, $1+y+y=11$
$\Rightarrow 2y+1=11$
$2y=11-1=10$
$\therefore y=\frac{10}{2}=5$
Substituting $y=5$ in (3) we get, $x=1+5=6$
$\therefore$ The number is $10x+y=10\left(6\right)+5=65$