Question

Ones digit of a 2 digit no. Is twice the tens digit when the no. Formed by reversing the digit is added to the original no. The sum is 99 find the original no

Answer 1

Step-by-step explanation:

let, the tens place digit of a number is x

then Ones place number is 2x

the first number is x*10+2x*1=12x

when digits are reversed the number would

10*2x+x=21x

A/Q,12x+21x=99

33x=99

x=99/33

x=3

there fore digits are x=3,2x=6

so numbers are 63,36.

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