In A,B,C are having some marbles with each of them. A has given 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 startinga. 20 b. 30 c. 32

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indu

  • Jul 25th, 2006
 

ans-- 1.b is

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Chimuelo

  • Apr 19th, 2007
 

First, you have
A : x
B : y
C : z

After x gives, you have
A : x-y-z
B : 2y
C : 2z

After y gives, you have
A : 2(x-y-z)
B : -x+3y
C : 3z

After z gives, you have
A : 4(x-y-z)
B : 2(-x+3y)
C : -x-y+5z

Therefore, the answers are:

1. -2x+6y
2. 39

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Jekin

  • Jun 30th, 2007
 

first
A: x
B: y
C: z

second
A: x-y-z

B: 2y

C: 2z


third

A: 2(x-y-z)

B: 2y-(x-y-z)-2z=3y-x-z

C: 4z


forth

A: 4(x-y-z)

B: 6y-2x-2z

C: 4z-2(x-y-z)-(3y-x-z)=7z-x-y

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