Question

sum of the digit of a two digit number is 9 and the number itself is equal to twice the sum of its digit find the number​

Answer 1

Answer:

The number is 18

Step-by-step explanation:

Given :

The sum of the digits of a two digit number is 9, the number itself is twice the sum of its digit.

To Find :

The number whose sum of digit is 9 and the number itself is twice the sum of its digit.

Solution :

Let the digits be x (in tens digit) and y (in ones digit)

Then, the digits in expanded form would be 10x + y , (E.g.,like 48 = 40 + 8)

The sum of the digits of this number is 9

⇒ x + y = 9   ...(i)

The number itself is twice the sum of the digits,.

⇒ 10x + y = 2 × 9

⇒ 10x + y = 18  ....(ii)

By subtracting (i) from (ii),

We get,

(10x + y) - (x + y) = 18 - 9

9x = 9

⇒ x = 1.

By substituting the value of x in (i),

We get,

1 + y = 9

y = 9 - 1

⇒ y = 8.

∴ The number is (10x + y) = 10(1) + (8) = 10 + 8 = 18

Answer 2

Answer:

The number whose sum of digit is 9 and the number itself is twice the sum of its digit. The number itself is twice the sum of the digits,. ⇒ x = 1. ⇒ y = 8