Question

The ones digitof a 2-digit number is twice the tens digit . When the number formed by reversing the digit is added to the original number,the sum is 99 . Find the original number

Answer 1

suppose
the ones digit is b
then A/q tens digit will be 10×b/2
therefore, the tens digit will be 10×b/2+b
=5b+b
=6b
Now
by reversing these numbers
10b+b/2+6b=99
16b + b/2= 99
(32b+b) / 2 = 99
33 b / 2 = 99
33b = 99 × 2
b= 99×2 / 33
b =3×2
b= 6
therefore,
the original number = 6b = 6×6=36

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