the ones digit of a two 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 unit digit be y & tens digit be x
Y=2x
After reversing digit the number will be10y+X
10y +x+10x+y=99
11X+11y =99
X+y=9
X+2X=9(putting the value of y=2x)
3x=9
X=3
y=6
The number will be 10x+y = 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