PK 2015 Q13

From the first picture, we can see:
\[\begin{array}{lrcll} \text{From the right scale:} & Z & > & Y & \quad (1) \\ \text{From the left scale:} & X & > & W & \quad (2) \\ \text{From the large scale:} & Y + Z & > & W + X & \quad (3) \end{array}\]
It follows from (1), (2) and (3) that \(Z + Z > Y + Z > W + X > W + W\), so \(2Z > 2W\), and hence \(Z > W\) … (4).
We can show that most swaps give a contradiction of these inequalities:
Firstly, suppose that weight \(Z\) doesn’t move. Then there are three possible swaps:
\(X\) and \(Y\) swap: then in the second picture we must have \(Z < X\) and \(Y < W\), which add to give \(Y + Z < W + X\), contradicting (3).
\(Y\) and \(W\) swap: then we would have \(Z < W\), contradicting (4).
\(X\) and \(W\) swap: then we would have \(X < W\), contradicting (2).
Hence \(Z\) must swap, which can happen in three ways:
\(Z\) and \(Y\) swap: then we would have \(Z < Y\), contradicting (1).
\(Z\) and \(X\) swap: then we would have \(Z < W\), contradicting (4).
\(Z\) and \(W\) swap: This is the only possibility left, and it can work when the weights \(W\), \(X\), \(Y\), \(Z\) are 1, 2, 3, 4. The swap would give 4, 2, 3, 1 which matches the picture on the right. However, this does depend on the values of \(W\), \(X\), \(Y\), \(Z\); it would not work for 1, 5, 3, 4.