Question

the ones digit of a two digit no. is twice thetens digit.when the no. formed by reversing the digita is added to the original no,the sum is 99. findthe original number

Answer 1

Let x be d ones digit of the original No. And y be the tens digit.
Then 10y+x is the original no.
Then after reversing digits, 10x+y becoms the no.
So, adding the original n new no. , we get

(10x+y)+(10y+x)=99
11x+11y=99
11(x+y)=99
X+y=9

Bcoz d ones digit is twice d tens digit, and their sum is 9,
The only possible value can be3 n 6.
So, the no. Is 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