Question 158698


    question 1:

    let the first number = x

   so, the second number = x + 3

  and the third number = 4x


    sum of all the numbers = 33
   
  so, x +(x+3) + 4x = 33

 or,   x+ x+ 3 + 4x = 33

 or,   6x + 3 = 33

  or,  6x = 33-3

        6x = 30

     so, x = 30/6 

      so, x= 5

      so, x+ 3 
         = 5+3 = 8


         4x = 4*5 = 20

  check :

  replacing x by 5, we get :

         5 + 8 + 20 = 33

so, the three numbers are  : 5, 8 and 20





 question 2:

   let the first number = n

   so, the second number is n +4

   the third number is 3n

    the sum of the numbers is 24

    so, n + (n+4) + 3n

      =  n + n + 4 + 3n

       = 5n + 4 

    so, 5n +4 = 24

         or, 5n = 24 - 4

         or, 5n = 20

         or,  n = 20/5
  
         0r, n = 4 

       so, n+4 = 4 + 4 = 8

          3n = 3 * 4 = 12

     check :

      replacing n by 4, we get :

        4+ (4+4) +(3*4)
  
        = 4 + 8 + 12

       = 24

so, the three numbers are 4, 8 and 12