If all the children sit in one of the rocker arms **How many girls should sit at the other end to maintain balance?**

To solve the puzzle we must assume that the distance of the children to the center of the rocker is not relevant. All boys are supposed to have the same "X" weight and all girls weigh the same "Y".

#### Solution

The illustration gives us a clear idea of the algebraic meaning of the two sides of the equation that we can obtain after months of complicated deductions. We will apply the first principle of algebra that tells us that the amounts added or subtracted on both sides of equality do not change the result so that we will solve the puzzle by the cancellation principle.

There are 5 children in one of the arms and 3 in the other, so we can eliminate 3 from each end while maintaining the balance of the rocker.

We have 3 girls left at one end and 6 at the other, so we will eliminate 3 girls from each end, in this way, we have 2 boys that balance the rocker with 3 girls.

Thus we conclude that two boys weigh the same as 3 girls.

Thus, **If we place the 8 boys at one end of the rocker, we will need 12 girls to balance it.**