Predicting a precipitate based on Ksp
To obtain $Q_\mathrm{sp}$ for

$$\ce{Ag2CrO4 &lt;=&gt; 2 Ag+ + CrO4^2-}$$

$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{CrO4^2-}]$$

$$Q_\mathrm{sp} = c(\ce{Ag+})^2\cdot c(\ce{CrO4^2-})\label{eqn:1}\tag{1}$$

you need to use new concentrations of ions in solution $c(\ce{Ag+})$ and $c(\ce{CrO4^2-})$, which can be found from the corresponding values for initial concentration $c_0$ and volume $V_0$:

$$
\begin{align}
c(\ce{Ag+}) &amp;= \frac{c_0(\ce{AgNO3})\cdot V_0(\ce{AgNO3})}{V_0(\ce{AgNO3}) + V_0(\ce{K2CrO4})} \\
  &amp;= \frac{\pu{0.10 mol L-1} × \pu{105 mL}}{\pu{105 mL} + \pu{125 mL}}\\
  &amp;= \pu{4.57E-2 mol L-1} \label{eqn:2}\tag{2}
\end{align}
$$

$$
\begin{align}
c(\ce{CrO4^2-}) &amp;= \frac{c_0(\ce{K2CrO4})\cdot V_0(\ce{K2CrO4})}{V_0(\ce{AgNO3}) + V_0(\ce{K2CrO4})} \\
  &amp;= \frac{\pu{0.35 mol L-1} × \pu{125 mL}}{\pu{105 mL} + \pu{125 mL}} \\
  &amp;= \pu{1.90E-1 mol L-1} \label{eqn:3}\tag{3}
\end{align}
$$

Finally, plug in the concentrations of ions from \eqref{eqn:2} and \eqref{eqn:3} in \eqref{eqn:1} and you find a value for $Q_\mathrm{sp},$ which indeed exceeds $K_\mathrm{sp}$:

$$Q_\mathrm{sp} = (\pu{4.57E-2 mol L-1})^2 × \pu{1.90E-1 mol L-1} = \pu{4.00E-4 mol3 L-3}$$