**Question:**

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

**Solution:**

Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$.

The sum of the digits of the number is 5 . Thus, we have $x+y=5$

After interchanging the digits, the number becomes $10 x+y$.

The number obtained by interchanging the digits is greater by 9 from the original number. Thus, we have

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

$\Rightarrow 10 x+y-10 y-x=9$

$\Rightarrow 9 x-9 y=9$

$\Rightarrow 9(x-y)=9$

$\Rightarrow x-y=\frac{9}{9}$

$\Rightarrow x-y=1$

So, we have two equations

$x+y=5$

$x-y=1$

Here *x* and *y* are unknowns. We have to solve the above equations for *x* and *y*.

Adding the two equations, we have

$(x+y)+(x-y)=5+1$

$\Rightarrow x+y+x-y=6$

$\Rightarrow 2 x=6$

$\Rightarrow x=\frac{6}{2}$

$\Rightarrow x=3$

Substituting the value of *x *in the first equation, we have

$3+y=5$

$\Rightarrow y=5-3$

$\Rightarrow y=2$

Hence, the number is $10 \times 2+3=23$.