The sum of digits of a two-digit number is 15. The number obtained by reversing the order of digits of the given number exceeds the given number by 9. Find the given number.

Given :

Sum of the digits of a two-digit number $= 15$.

When we interchange the digits, the resulting new number exceeds the original number by $9$.

To do :

We have to find the given number.

Solution :

 Let the two-digit number be $10x+y$.

$x + y = 15$

$x=15-y$.....(i)

The number formed on reversing the digits is $10y+x$.

Therefore,

$10y+x = (10x+y)+9$

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

$9(y-x) = 9$

$y-x = 1$

$y-(15-y) = 1$    (From (i))

$y+y = 1+15$

$2y = 16$

$y = 8$

This implies,

$x = 15-8 = 7$

The original number is $10(7)+8 = 70+8 = 78$.

The original number is 78.