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. 20 b. 30 c. 32
3.

• 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
  
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• Start it from the end
  A B C
  24 24 24
  12 12 48
  6 42 24
  39 21 12
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