Question

the sum of digits of a two digit number is 5. the number obtained by reversing the order of digit diminished the original number by 9 .find the number.

Answer 1

a+b = 5 , (10a+b)-(10b+a)= 9 now solve yourself please mark brainliest

Answer 2

Here's your answer !

Let the digit in the units place be x
Digit in the tens place = 5 - x

The original number is

10(5 - x ) + x = 50 - 10x + x = 50 - 9x_ _ _ _ (i)

Given,

the number obtained by reversing the order of digit diminished the original number by 9


So , by interchanging the digits , we have ,


Digit in units place = 5 - x
Digit in tens place = x

The new number is 10(x) + 5 - x = 10x + 5 - x
= 9x + 5 _ _ _ (ii)

By balancing the equation,
we have,


50 - 9x = 9x + 5 + 9
50 - 9x = 9x + 14
50 - 14 = 9x + 9x
36 = 18x

$x = \frac{36}{18}$

x = 2


So,

Digit in units place = x = 2
Digit in tens place = 5 - x = 5 - 2 = 3


Thus ,

The original number is 32