Assume the digit in ones place as x and the digit in tens place is y
The original number is (10y + x)
Number obtained by reversing the digits = (10x + y)
Now check for the condition
Given that sum of digits of the number is 9
That is x + y = 9 → (1)
Second condition is, number obtained by interchanging the digits is greater than the original number by 27
That is (10x + y) = (10y + x) 27
⇒ 10x + y – 10y – x = 27
⇒ 9x – 9y = 27
∴ x – y = 3 → (2)
Add (1) and (2), we get
x + y = 9
x – y = 3
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2x = 12
∴ x = 6
Put x = 6 in (1), we get
6 + y = 9
⇒ y = 9 – 6 = 3
The original number = 10y + x = 10(3) + 6 = 36