Question 28142
The sum of the digits of a two-digit number is 9. If the digits are reversed, the new number is 27 less than the original. Find the original number.

Let the digit in the units place be u and
let the digit in the tens place be t
Then the two digit  number is givn by (10t+u)----(*)
By data the sum of the digits of the two digit  number is 9
That is (u+t) = 9----(1)
Now if the digits are reversed, that is u and t switch places then now t is the unit place digit and u is the tenth place digit
This reversed number is  (10u+t)and it is given to be 27
That is (10u+t) = 27 ----(2)
Putting (1) in (2) (using t = 9-u )
  10u +(9-u) = 27
  (10u-u)= 27-9
    9u = 18
   u = 18/9 = 2
u= 2 in (1): u+t = 9 implies t = 7
Therefore the original number is got by putting t=7 and u =2 in (*)
(10t+u) = 10X7 + 2 = 70+2 = 72
Answer: The required number is 72
Verification: IF we add up the digits in 72 we get 9 and 72 when reversed is 27.
Therefore our answer72 is correct