Question

a number contest of the two digits whose sum is 8 if 18 is added to the number its digits are reserved find the number

Answer 1

let the 2 numbers be 'x' and 'y'
it is given that the sum of its digits is 8
therefore x + y=8,this implies x=8 - y
according to the question the equation is as follows:
     
          10x + y +18=10y+x
         10(8 - y)+y+18=10y +(8-y)   [because x=8-y]
           80-10y+y+18=9y+8
            80-9y+18=9y+8
            80-8+18=9y+9y
             72+18=18y
            90=18y
  therefore,y=5
    now we know that x=8-y
   so by this equation  x=8-5
                                   x=3
now,x=3 and y=5
therefore the number is 10x +y=10(3) + 5
    so the number is 35....!!!
    hope this answer is helpful to you.....

Answer 2

Let the digit of tens place be x
Of units place be y
The number is 10x+y
According to the question,
x+y=8
10x+y+18=10y+x(as the digits are reversed)
=>9x+18=9y
Divide by 9
x+2=y
=>2x+2=8
=>x=3
y=x+2=5
Therefore the number is 35