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.
Answers
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:
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