Question

If the sum of the digits of the two digit number is 9. If we interchange the digits the resulting number is 27 greater than the original number.Find the original number.​

Answer 1

Answer:

Let the digits at tens place and ones place: x and 9−x respectively.

∴ original number =10x+(9−x)

=9x+9

Now Interchange the digits: Digit at ones place and tens place: x and 9−x respectively.

∴ New number: 10(9−x)+x

=90−10x+x

=90−9x

AS per the question

New number = Original number +27

90−9x=9x+9+27

90−9x=9x+36

18x=54

x=

18

54

x=3

Digit at tens place ⇒3 and one's place : 6

∴ Two digit number: 36

Answer 2

Answer:

$\huge\green{\boxed{\sf Answer}}$

Given

The sum of the two digits = 9

On interchanging the digits, the resulting new number is greater than the original number by 27.

Let us assume the digit of units place = x

Then the digit of tens place will be = 9–x

Thus the two-digit number is 109–x + x

Let us reverse the digit

the number becomes 10x + 9–x

As per the given condition

10x + 9–x = 109–x + x + 27

⇒ 9x + 9 = 90 – 10x + x + 27

⇒ 9x + 9 = 117 – 9x

On rearranging the terms we get,

⇒ 18x = 108

⇒ x = 6

So the digit in units place = 6

Digit in tens place is

⇒ 9 – x

⇒ 9 – 6

⇒ 3

Hence the number is 36