Hi Cathey. You can use algebra to solve this problem.
Let M = the total number of marbles of all the children combined.
Then the first child has (1/2)M  4 marbles.
The second child has (1/5)M + 6 marbles.
The third child has (1/3)( (1/2)M  4) marbles.
The fourth child has ( (1/5)M + 6 )  1 marbles.
We can add up these totals in two ways. The simple way is just to
say that the total is M, as we decided in the beginning. The other
way is to add up each child's marble count. Since these two
quantities must be the same, they equal each other. So:
M = ( (1/2)M  4 ) + ( (1/5)M + 6) + ( (1/3) ( ( 1/2) M  4) ) + ((1/5)M + 6 )  1.
If you simplify this equation, you can "solve for M" and get the answer to your question.
Hope this helps,
Stephen La Rocque.
