Question

A number consists of two digits whose sum is $9$. The number formed by reversing the digits exceeds twice the original number by $18$. Find the original number.

Answer 1

Let the ten's digit be $x$.

Let the unit's digit be $y$.

So, $x+y=9$

So, $y=9-x$

The actual number is $=10x+y$

According to the question,

$10y+x-2(10x+y)=18$

$10y+x-20x-2y=18$

$8y-19x=18$

$8(9-x)-19x=18$ $[$ Substituting the value of $y]$

$72-8x-19x=18$

$72-18-27x=0$

$54=27x$

$x=2$

So,

$y=9-2$

$y=7$

So, the original number is $2\times 10+7=27$

Hence, this is the answer.