KK the question is solved by basic algebra as some of the earlier posters have demonstrated, though I suspect that an eight year old would be expected to solve it using substitution.

In both cases the key is to start with the number of coins in the second pile as the contents of the other piles (and by implication the total number of coins) are derived from the second pile.

So, algebraically if there are x coins in pile 2: then (x+4)+x+(x+1) +2x=20

Ie 5x+5=20: 5x=15 and x=3, which determines 7 3 4 6

By substitution:

Pile 2=1, hence pile 1=5, pile 3=2, pile 4=2...total 10 - no!

Pile 2=2, hence pile 1=6, pile 3=3, pile 4=4...total 15 - getting there!

Pile 2=3, hence pile 1=7, pile 3=4, pile 4=6...total 20 - result!

Try this: A salesman is entitled to 10% commission on his sales after his commission has been charged. If his commission is 18 pounds, what are his sales?

Edited: 28/01/2013 at 18:24