Question

the ones digit of a 2 digit number is twice the tense digit when the number formed by reversing the digit is added to the orignal number. the sum is 99 . find the original number

Answer 1

Ten’s digit: x
Ones’s digit: 2x
No: 10(x) + 2x
Reversed no: 10(2x) + x
According to the question:
10x + 2x + 20x + x = 99
33x = 99
x = 99/33 = 3
No: 10(3) + 2(3) = 30 + 6 = 36

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