We had this solution sent in from Emma at Kelso State School

1. You could put two on 3$\frac{1}{2}$.
2. You could also put one on the $\frac{1}{2}$ and the 6$\frac{1}{2}$.

Summer and Yesha at Valence Primary School sent in the following:

7 = 0.5, 6.5
= 1.5, 5.5
= 2.5, 3.5

7 = 14 halves

Boyang from Mountfields Lodge and Caroline from St Jude's C of E School looked at this problem very logically. To balance 7 on the left using two "half" weights on the right, Boyang told us:

The two numbers on the right hand side could be:

$\frac{1}{2}$ + $6\frac{1}{2}$
$1\frac{1}{2}$ + $5\frac{1}{2}$
$2\frac{1}{2}$ + $4\frac{1}{2}$
$3\frac{1}{2}$ + $3\frac{1}{2}$
$4\frac{1}{2}$ + $2\frac{1}{2}$
$5\frac{1}{2}$ + $1\frac{1}{2}$
and finally, $6\frac{1}{2}$ + $\frac{1}{2}$.

I wonder whether the first solution is different from the last one? What do you think?

Caroline wasn't sure whether you are allowed two weights on the same hook, but I think that's fine.

Nobody has looked at hanging three weights on the right-hand side, or having quarter divisions on the balance. If you do investigate these situations, let us know! Please don't worry that your solution is not "complete" - we'd like to hear about anything you have tried.  Teachers - you might like to send a summary of your children's work.