In A,B,C are having some marbles with each of them. A has giben B and C the same number of marbles they already have to each of them. then, B gave C and A the same no. of marbles they have, then C gave A and B the same no. of marbles they have. At the end A,B,and C have equal no. of marbles. (i) If x,y,z are the marbles initially with A,B,C respectively. then the no of marbles B have at the end (a) 2(x-y-z) (b) 4(x-y-z) etc. (ii)If the total no. of marbles are 72, then the no. of marbles with A at the starting

• A B C
  x y z (Initially)
  x-y-z 2y 2z (After A gave)
  2(x-y-z) 3y-x-z 4z (After B gave)
  4(x-y-z) 2(3y-x-z) 7z-x-y (After C gave)
  
  So B has 2(3y-x-z) at the end
  
  x+y+z=72
  4(x-y-z)=24 (ie72/3)
  
  solving
  we get x=39
  so A had 39 at the beginning
  
• 10 years agoHelpfull: Yes(17) No(3)
• Here total number of marbles are 72 we know that a, b, C have equal no of marbles at the end, so just find out how many marbles A has at the end x+y+z=72
  A's transaction
  A------x-y-z(because it is loosing y&z)
  B------(y+y) - - - - - A gave same no Mar what B have already so it is 2y
  Similarly C has 2z
  B's transaction
  A_______X-y-z+(x-y-z)
  B____2y-(x-y-z)-2z
  3y-x-z
  C_____2z+2z
  C transaction
  A_____4(x-y-z)
  B_____2(3y-x-z)
  C_____4z-2(x-y-z)-(3y-x-z)
  7z-x-y
  Each have 24 marbles (72/3)
  Finally A has
  4(x-y-z)
  4(x-y-z)=24_____x-y-z=6
  X+y+z=72
  2x=78
  ******x=39******
• 7 years agoHelpfull: Yes(2) No(0)
• Let they have x, y and z respectively.
  x-y-z , 2y, 2z [1st transaction]
  2(x-y-z), 2y-(x-y-z)-(2z), 4z [2nd transaction]
  2(x-y-z), 3y-x-z, 4z [2nd transaction]
  4(x-y-z), 2(3y-x-z), 4z-(2x-2y-2z)- (3y-x-z) [3rd transaction]
  4(x-y-z), 2(3y-x-z), 7z-x-y [3rd transaction]
  
  B has 2(3y-x-z) at the end of the transaction.
• 7 years agoHelpfull: Yes(0) No(0)