Question

Sum of the digits of a two digit number is 9. When we interchange the digits the new number is 27 greater than the earlier number. Find the number.

Answer 1

Answer:

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 10(9 – x) + x

Let us reverse the digit

the number becomes 10x + (9 – x)

As per the given condition

10x + (9 – x) = 10(9 – 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

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