Question

The sum of the digit of a two-digit number is 11 .The number is less than the number obtained by interchanging the digit by 9 . Find the number.​

Answer 1

The number is 65.

Explanation:

Let, the the tenth digit be, 'y'.

The sum of the digits is 11.

∴ The unit's digit = (11-y)

∴ The number = (y×10)+{(11-y)×1}

                       = 10y+11-y

                       = 9y+11

By interchanging the digits we get the new number as_

{(11-y)×10}+(y×1)

= 110-10y+y

= 110-9y

According to the problem,

(9y+11) - (110-9y) = 9

⇒ 9y+11 -110 +9y = 9

⇒ 18y-99 = 9

⇒ 9(2y-11) = 9

⇒ 2y-11 = 9/9

⇒2y-11 = 1

⇒2y = 11+1

⇒ y = 12/2

⇒ y = 6

Hence, the tenth digit is 6.

∴ The unit's digit = (11-6)

                           = 5

∴ The number = (6×10)+(5×1)

                          = 60+5

                         = 65

So, the number is 65.

Check whether the answer is correct or wrong:

We get,

Tenth digit = 6

Unit's digit = 5

The sum of the digits = (6+5)

                                   = 11

By interchanging the digits we get the new number as_

(5×10)+(6×1)

= 50+6

= 56

The new number is smaller than the previous number by_

(65-56)

= 9