= Solution
Use $K^-=s\bar u$ and $p=uud$. In the first reaction a <proton> $u$ can join the incoming $\bar u$ to form the $u\bar u$ component of the <neutral pion>, leaving $uds$. In the second, a $u\bar u$ pair can annihilate into a <gluon> which creates $s\bar s$; the remaining $u$ joins $\bar s$ in the outgoing <kaon>, leaving $dss$. Thus
$$
\boxed{X^0=uds,\qquad Y^-=dss.}
$$
<baryon number> is one in both cases. Strong <strangeness> conservation provides a separate check: the first <baryon> has <strangeness> $-1$; the second has <strangeness> $-2$ because the outgoing $K^+$ carries $+1$. The <Feynman diagrams> show representative <quark-line flavour rearrangement> and pair-annihilation topologies. Bound <hadrons> require nonperturbative binding, so a single drawn <gluon> is not an exclusive cross-section calculation.
\Image[/past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2014/paper-1-quark-flow.png]
{title=Original representative quark-line diagrams: flavour rearrangement and u-antiu annihilation into an s-antis pair}
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