If A,B,C are having some marbles with each of them. A has given B and C the same number of marbles each of them already have. Then, B gave C and A the same number of marbles they already have. Then C gave A and B the same number of marbles they already have. At the end A, B, and C have equal number of marbles. If x,y,z are the marbles initially with A,B,C respectively. Then the number of marbles B have at the end

(a) 2(x-y-z)
(b) 4(x-y-z)
(c) 2(3y-x-z)
(d) (x+y-z)

• 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
• 12 years agoHelpfull: Yes(62) No(2)
• (c) 2(3y-x-z)
• 14 years agoHelpfull: Yes(48) No(7)
• (c) 2(3y-x-z)
• 14 years agoHelpfull: Yes(22) No(5)
• (b) 4(x-y-z)
• 14 years agoHelpfull: Yes(11) No(19)
• A B C
  
  INITIALLY X Y Z
  
  A gives B & C X-Y-Z 2Y 2Z
  
  B gives C & A 2(X-Y-Z) 2Y-(X-Y-Z)-2Z 4Z
  
  C gives A & B 4(X-Y-Z) 2(2Y-(X-Y-Z)-2Z) {4Z-2(X-Y-Z)
  -(2Y-(X-Y-Z)-2Z)}
  
  
  
  NOW SOLVE THE ABOVE EQN TO GET THE ANSWER
• 12 years agoHelpfull: Yes(11) No(4)
• c
• 14 years agoHelpfull: Yes(9) No(4)
• going in reverse order
  A B C
  a a a
  a/2 a/2 2a
  a/4 7a/4 a
  13a/8 7a/8 4a/8
  which is x y z
  so option b is rite
• 12 years agoHelpfull: Yes(4) No(10)
• 6y-2z-2x is ans
• 9 years agoHelpfull: Yes(1) No(1)
• 6y-2z-2x is ans
• 9 years agoHelpfull: Yes(0) No(1)
• Please explain it clearly .how we get those equations
• 8 years agoHelpfull: Yes(0) No(0)