Question

the ones digit of a 2-digit number is twice the tens digit. when the number formed by reversing the digits is added to the original number, the sum is 99. find the original number.
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Answer 1

Answer:

Let the tens digits =x

so, ones digits = 2x

number= 10x+2x=12x

reverse number= 10{2x}+x=20x+x=21x

it is given that

12x+21x=99

or 33x= 99

so,x=3

original number =10x+2x

30+6

36 answer

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