rakesh0814069

 Profile Answers by rakesh0814069 

• Aug 9th, 2010

 

 A to B y
A to C z
A has x-(y+z)
B has 2y
C has 2z
b to A x-(y+z)
b to c 2z
B has 2y-(2z+x-y-z)
  2y-z-x+y
3y-(x+z)

rakesh0814069

 Profile Answers by rakesh0814069 

• Aug 9th, 2010

 

step 1 a to
b=y
c=z
end a=x-(y+z)
      b=2y c=2z
step 2 B to
c=2z
a=x-(y+z)
end
a=2(x-y-z)
b=2y-(2z+x-y-z)=3y-z-x
c=4z
step 3  C to
a=2(x-y-z)
b=3y-z-x
now b has=2(3y-z-x)

Md Parvez

 Profile Answers by Md Parvez 

• Mar 16th, 2011

 

first a,b and c have same number of marbles.
a->x
b->y
c->z
now a gave b and c so
a->-x
b->y+y
c->z+z
now b gave a and c
a->-x+x
b->0
c->z+z+z
now c gave a and b
a->-x+x+x
b->0+y
c->z


it means a have x, b have y and s have z.same as they had initially.