In an examination , the number of those that passed and the number of those that failed were in the ratio 3:1.Had 8 more appeared, and 6 less passed, the of passes to failures would have been 2:1.Find how many appeared?

let students passed  = x
    students failed    =  y 
    Total students  = x + y 

   x  / y =  3 / 1    (given as ratio)
  x  =  3y
 x - 3y  = 0
 
NOW 8 more students appeared and 6 less passed 
 Total students = x + y + 8
  students passed = x - 6
 
SO students failed = x+y +8 -(x-6)
                              =  x+ y +8 - x +6 
                              =  y + 14

       (x - 6 ) / (y + 14) =  2 / 1
       x - 6 = 2y + 28 
      x - 2y  =  34
 BY ELIMINATION METHOD 

  x - 3y = 0
 x  -  2y = 34
      y  =  34

x = 3*34 
  =  102
Total students appeared = 102 + 34 = 136
 
Hope this will help