Question

The ones digit of a two 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

let u=ones digit
let t=tens digit
u=2t
10t+u=original number
10u+t=reversed number
10t+u+10u+t=99
10t+2t+20t+t=99
33t=99
t=3
u=6
original number=10t+u=60+3=63 

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