SOLUTION: Hi there were 900 sweets in boxes a b c.18 sweets were transfered from a to b.5 sweets were transferred from b to c.1/3 of the sweets in c were then transferred to a.there were an

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Question 1098822: Hi
there were 900 sweets in boxes a b c.18 sweets were transfered from a to b.5 sweets were transferred from b to c.1/3 of the sweets in c were then transferred to a.there were an equal number of sweets in all 3 boxes. How many sweets were in each box at first.
thanks

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
If initial amounts in each box given variables a, b, c,
and if the description is followed exactly, this system of equations can be made:



Algebra steps leads to .
.
.

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
The condition says that after all exchanges there were an equal number of sweets in all 3 boxes.

 = 300, which means that finally there were 300 sweets in each box.


Further, the condition says


a - 18 +  = 300,    (1)

b + 18 - 5 = 300,           (2)

(c+5) -  = 300.     (3)


From (3),   = 300  ====>  c+5 =  =  = 450  ====>  c = 450-5 = 445.

From (2),  b = 300 + 5 - 18 = 287.

From (1), a = 300 -  + 18 = 300 - 150 + 18 = 168.



Answer.  Originally, there were 168 sweets in A, 287 sweets in B  and 445 sweets in C.

Solved.


      There is NO NEED to solve system in 3 equations in 3 unknowns.

      This problem is for young students (5 - 6 grades) who have no any notion on systems of equations.

      The solution and the approach by @josgarithmetic go entirely and totally out of the target.


      For your safety, simply ignore his writing.



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