Question

the ones digit of a two digit number is twice the tens digit and the number formed by reversing the digit is adding to the original number the sum is 99 find the original number

Answer 1

LET tens digit number=x
Ones digit number=2x

As tens digit number means 10× given value

So Number=10×x+2x
Number=10x+2x

Reverse number=10×2x+x
Reverse number=20x+x

Their sum=99

A/Q
10x+2x+20x+x=99
33x=99
x=99/33
x=3
Tens digit =3
Ones digit=2x=2×3=6

Number=10×x+2x
10×3+2×3
36

Original Number=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