Question

the ones digit of a two digit number is twice the tens digit when the number formed by reversing the digits are is added to the original number the sum is 99 find the original number

Answer 1

Let tens digit is a

Ones digit = 2a

Original no = 10a + 2a = 12a

Reversed no = 20a + a = 21a

Eq. =

12a + 21a = 99

33a = 99

a = 99 / 33

a = 3

Original number = 36

HOPEFULLY THIS WILL HELP YOU

MARK AS BRAINLIEST....

Answer 2

Answer:

Let the tens digit be y and the ones digit be x.

The original number = 10y + x

The reverse number = 10x + y

It is given that ones digit is twice the tens digit :]

➳ x = 2y ............[Equation (i)]

According to question now,

➳ 10x + y + 10y + x = 99

➳ 11x + 11y = 99

➳ 11 (x + y) = 99

➳ x + y = 99/11

➳ x + y = 9

➳ y = 9 - x.........[Equation (ii)]

Now, Substituting equation (ii) in equation (i) we get :

➳ x = 2 (9 - x)

➳ x = 18 - 2x

➳ 3x = 18

➳ x = 18/3

➳ x = 6

Putting x = 6 in equation (ii) we get :

➳ y = 9 - x

➳ y = 9 - 6

➳ y = 3

Therefore,

The original number = 10y + x = 10(3) + 6 = 30 + 6 = 36