Question

the sum of the digits of a two digit number is 9 .if this number exceeds a number formed by reversing its digits by 45.find the number

Answer 1

let digit at ones place be x
digit at tens place be 10-x
so the no. would be 10 (10-x)+x= 100-99x

another no
let digits at ones place be 10-x
digits at tens place be x
so another no. be 10×x+(10-x)=9x+10+1

Eq.
{100-99x} - {9x+10}=45
90 - 90x= 45
-90x = 45-90
-90x= -45
x=45

so the no. are
ones digit=1/2=0.5
tens digit= 10- 0.5 = 9.5
so the no = 10×9.5+0.5=95.5

Another no.
ones digit = 10-0.5=9.5
tens digit = 0.5×10=5
so the no =59.5+1=60.5

verification:-
difference=45
95.5-60.5=45