The sum of two digit number …

Let the ten's and unit's digits of the required number be x and y respectively.
Then, the number = (10x + y).
The number obtained on reversing the digits = (10y + x).
As per given condition
The sum of a two-digit number and the number obtained by reversing the order of its digits is 99.
{tex}\therefore{/tex} (10y + x) + (10x + y) = 99
{tex}\Rightarrow{/tex} 11(x + y) = 99
{tex}\Rightarrow{/tex} x + y = 9.
The digits differ by 3
So, (x - y) = ±3.
Thus, we have
x + y = 9 ........ (i)
x - y = 3 .......... (ii)
or x + y = 9 ......... (iii)
x - y = -3 ............ (iv)
From (i) and (ii), we get x = 6, y = 3.
From (iii) and (iv), we get x = 3, y = 6.
Hence, the required number is 63 or 36.

first no.=10x+ y reversing no.=10y+x A.T.Q (10x+y)+(10y+x)=99 11x+11y=99 x+y=99/11 x+y= 9 (1) If x-y=3 (2) then solve by elimiation x+y=9 x-y=3 - + - 2y=6 y=6/2 y= 3 putting value of y=3 in (2) x- 3=3 x=3+3=6 . . . the number= 63 If y-x=3 (3) by elimination x+y=9/-x+y= 3 2y= 12 y= 12/2=6 putting y=6 in(1) x+6=9 x=9-6=3 . . . the number=36 hence, the sum of these numbers are. 63 + 36= 99 and. 36+63=99. Ans.