Question

the ones digits 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 ONES PLACE DIGIT BE 2x THEREFORE TENS PLACE DIGIT WILL BE x ,
HENCE THE NUMBER WILL BE 10×x + 2x = 12 x. ..........1
After reversing the number we get the number 10×2x +x = 21x. ..........2
according to the question add eq. 1 and eq. 2
we have, 12x + 21x = 99
33x =99
X= 3
henceforth the number is 36.

Answer 2

let the ones digit be x
and tens digit be y
so the 2 digit no. will be 10×y+x = 10y+x
ATQ
when 2 digit no. reverse it become 10x+y
according to condition
10y+x +10x+y = 99
11y + 11x = 99
divided by 11
y +x = 9 ........(1)
NOW, IT IS GIVEN THAT ONES DIGIT IS TWICE THE TENS DIGIT
x = 2y
putting value of x in eqn (1)
y + 2y = 9
3y = 9
y = 3

the two digit no. is 10y+x
=10y+2y =12y
= 12×3 = 36

HOPE THIS IS HELPFUL !