Question 1063056
The sum of digits of a two-digit number is 9. On interchanging the order of the digits,the number becomes 27 more then the original number.find the number

Let a be the 10's digit
     and b be the 1's digit

The number is  10a+b

(1)  a+b = 9 (given)
(2)  10b+a = (10a+b)+27    (interchanging digits results in a number that is 27 greater than original)

(1) ==>  a=9-b
Substitute for '9-b' for 'a' in (2), that will give us an equation with just 'b' and we can solve for 'b'
 10b + (9-b) = (10(9-b) + b) + 27
   9b + 9 = 90 - 10b + b + 27
   9b + 9 =  117 - 9b
         18b = 108
             b = 108/18 = 6

b=6 ==>  a=9-6=3  (from (1))

The number is {{{highlight(36)}}}
--                              
Check:
     3+6  = 9 (ok)
      36 + 27 = 63  (= the interchanged version of 36, ok)