The ones digit of 2-digit number is twice the ten digits .when the nymber formed by reversing the digits is added to the original number, the sum is 99. find the original number
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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
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