The sum of two digit number and the number obtained by reversing the order of its digit is 99.if the digits differ by 3,find the number?

Let the ten's place be x and units place be y.

so the original number becomes 10x+y

so

 the number obtained by reversing the digits is 10y+x

ATQ

(10x+y)+(10y+x)=99

11x+11y=99

x+y=11....(i)

also

ATQ,

digits differ by 3

so

either x-y=3 or x-y=-3

lets take

x-y=3..(ii)

so , solving  (i) and (ii)

2x=14

so

x=7

and y=4

so original number is 74

next

 lets take x-y=-3..(iii)

so,

solving (i) and (iii)

2x=8

so x=4

and

y=7

so the number is 47