A solution having composition $p$ (left side of graph) is cooled to just above the eutectic temperature (point $s$ is at $0.18$ $x_{\ce{Si}}$ and the corresponding tie line intersects the liquidus curve at $0.31$ $x_{\ce{Si}} $); calculate the composition of the solid that separates and that of the liquid that remains.
The liquid phase is obviously 31 % silicon and 69 % gold as given in the question. The solutions manual also gives the weight percentage of the solid phase. Using the lever rule,
$$ m_\text{solid}(0.18)=m_\text{liquid}(0.31-0.18)$$ $$ \frac{m_\text{solid}}{m_\text{liquid}}=\frac{0.13}{0.18} $$ $$ \frac{m_\text{solid}}{m_\text{solid}+m_\text{liquid}} = \frac{0.13}{0.13+0.18}=0.42$$
However, the solution manual states that the wt% of solid phase is 58 %, and that of the liquid phase is 42 %. Is the solutions manual wrong, or am I losing my mind?