Question

the sum of digits of a two digit number is 12.the number obtain by reversing order of digit is 18 more than original number.find the reverse number.

Answer 1

Let first no. be x
Other be y
Original no. =10x+y
On reversing no will be: 10y+x
According to question
1) x+y=12
2) 10y+x=10x+y+18
9y=9x+18
y=x+2
y-x=2
Add both equations
=> x+y+y-x=2+12
=> 2y=14
=> y=7

Put value of y in eq. 1
=> x+y=12
=> x+7=12
=> x=12-7
=> x=5

So original no. is 57
Then reversed no. is 75

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Answer 2

Let the digit at one's place be=x
And the digit at ten's place be=y
Therefore, the no. is =x+10y
Now, according to the question:-
x+y=12----(i)
No. Obtained by reversing order of digits =10x+y
Again according to the question :-
10x+y=18+x+10y
10x+y-18-x-10y=0
9x-9y-18=0
9x-9y=18
9(x-y)=18
x-y=18/9
x-y=2----(ii)
Adding (i) and (ii).We get:-
2x=14
x=14/2
x=7.
Subtracting (i) and (ii). We get :-
2y=10
y=10/2
y=5.
Therefore the reversed no. =10(7)+5=75