Question

sum of the digits of a two-digit number is 9 .When we interchange the digits,it is found that the resulting new number is greater than the original number by 27. What is the original number?

Answer 1

let the unit place digit = x and tenth place= y number = 10y+x revrsed no. = 10x+y according to first condition x+y = 9 (i)
according to second condition
10y+x = 10x+y+27
10y-y+x-10x = 27
9y-9x=27
y-x = 3
-x+y = 3 (ii)
on solving eq (i) and (ii) by elimination method
x+y =9
-x+y =3
2y = 12
y = 6
putting in eq (i)
x+6=9
x=9-6
x = 3
thus number = 10×6+3
63

Answer 2

Here is my answer.. I have explained in short by this attachment