In 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 endA. 2(x-y-z)B. 4(x-y-z)C. 2(3y-x-z)D. x + y-z

C

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