Question

Sum of the digits of a 2 - digit number is 9. When the digits are reversed,it is found that resulting number is greater than the original by 27 . Find the number

Answer 1

Solution -

Let the digit in ones place be x

So, the digit in tens place be 9 - x

Original no. = 10(9 - x) + 1(x)

= 90 - 10x + x

= 90 - 9x

New no. = 10(x) + 1(9 - x)

= 10x + 9 - x

= 9x + 9

According to Question,

New no. - Original no. = 27

⇒ (9x + 9) - (90 - 9x) = 27

⇒ 9x + 9 - 90 + 9x = 27

⇒ 18x - 81 = 27

⇒ 18x = 27 + 81 = 108

⇒ x = 108/18

⇒ x = 6

Measurements -

Original no. = 90 - 9x

= 90 - 9(6)

= 90 - 54  = 36

New no. = 63 [∵ By reversing the digit ]

∴ The required number is either 36 or 63

#Be Brainly

Answer 2

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 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