Question

Sum of the digits of a two digits number is $9$. the number obtained by reversing the digits exceeds given number by $27$. Find out the given number.

Answer 1

Let the two digit number be $10x+y$.

Given that the sum of the digits is $9$

$x+y=9$ $.......\left(1\right)$

Given that the number obtained by interchanging the digits exceeds the given number by $27$

$10y+x=10x+y+27$

$9x-9y=-27$

On taking $9$ as common

$x-y=-3$ $...........\left(2\right)$

Adding equation $\left(1\right)$ and $\left(2\right)$

$x+y=9$

$x-y=-3$

$2x=6$

$x=3$

$3+y=9$

$y=6$

Therefore, the number is $10x+y$ is $36.$

Hence, this is the answer.