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## 'Inside Outside' printed from http://nrich.maths.org/

A good start by Jia and Jeremy from Raffles Institution, Singapore.

To balance a bar balance, equal weights are required on each side of the pivot.

Taking $a$ as the distance of the $3$ weight from the pivot, and $b$ as the distance of the $2$ weight from the pivot. And using each column from the pivot as a distance of one, the weight on the left of the pivot has a moment of $40$.

So $3a+2b$ must equal (that is, balance) $40$.

Matching pairs of $a$ and $b$ could be:

$$a = 5 \quad\mbox{and}\quad b = 12.5$$

or

$$a = 4 \quad\mbox{and}\quad b = 14\;.$$

And some good algebra reasoning from Joan in Edinburgh.

If we want $a$ less than $b$ it would be good to find where $a = b$, that means that the two weights are together at the same place.

If $a = b$ then $a$ and $b$ have to be $8$ (from $5a = 40$ or $5b = 40$ whichever you prefer).

But we want $a$ less than $b$, so a can't be more than $8$.

The $3$ weight can go as near to the pivot as we like and $b$ will just have to get bigger to keep the balance. If the $3$ weight does go to zero $b$ will have to be $20$.

$20$ is the furthest out from the pivot that the $2$ weight goes if it keeps balanced with the $4$ weight on the other side.