Question

The o
es 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.​

Answer 1

Let the digit in the ones place be y.

Let the digit in the tens place be x.

The original number=(x *10)+y

Given, y=2x

On the reversing the digits, the number =(y*10)+x

Given,

(y*10) + x + (x*10) + y =99

Substituting y=2x,

(2x*10) + x + (x*10) + 2x = 99

20x + x + 10x + 2x =99

33x = 99

x = 3

Substituting x=3 in y=2x,

y=2*3=6

Therefore, the original number =(x*10)+y

=(3*10)+6

=36

The original number 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