A problem regarding solubility and concentration A solution is $\pu{0.10 M}$ $\ce{Ba(NO3)2}$ and $\pu{0.10 M}$ $\ce{Sr(NO3)2}$. If solid $\ce{Na2CrO4}$ is added to the solution, what is the $\ce{[Ba+]}$ when $\ce{SrCrO4}$ begins to precipitate?  $\mathrm{K_{sp}}=\pu{1.2 \times 10^{-10}} \text{ for } \ce{BaCrO4}$
$\mathrm{K_{sp}}=\pu{3.5 \times 10^{-5}} \text{ for } \ce{SrCrO4}$

The solution given in my textbook is:

$\mathrm{K_{sp1}}=\ce{[Sr^2+][CrO4^2-]}$ $$\ce{[CrO4^2-]}=\frac{3.5\times 10^{-5}}{0.1}=3.5\times 10^{-4} \pu{M}\tag{1}$$ $\mathrm{K_{sp2}}=\ce{[Ba^2+][CrO4^2-]}$ $\ce{[CrO4^2-]_{total}}\approx\ce{[CrO4^2-]}\text{ from } \ce{SrCrO4}$ $$\ce{[Ba^2+}]=\frac{1.2\times 10^{-10}}{3.5\times 10^{-4}}=3.4\times 10^{-7}\pu{M}$$

First of all, in equation $(1)$ why did they put $\ce{[Sr^2+]}$ same as the initial concentration (before the addition of $\ce{Na2CrO4}$)? When $\ce{Na2CrO4}$ is added,  $\ce{[Sr^2+]}$ will react with $\ce{CrO4^2-}$ (obtained from $\ce{Na2CrO4}$) to form precipitate. The concentration of $\ce{[Sr^2+]}$ will undoubtedly change until the equilibrium is established.
This given solution is totally confusing, I would appreciate if you could tell me the proper way of solving this problem. I tried this on my own, but after calculating the concentration of $\ce{CrO4^2-}$, I couldn't make sense out of this question.
It is supposed addition of solid $\ce{Na2CrO4}$ would cause negligible volume change, so we can afford to consider the same $[\ce{Sr^2+}]$ as the initial one.
The task is rather theoretical, supposing the time point when $\ce{[Sr^2+][CrO4^2-]}$ just reached $K_\mathrm{sp,\ce{SrCrO4}}$, while $K_\mathrm{sp,\ce{SrCrO4}}$ and $\ce{[Sr^2+]}$ are given.
The proper computation procedure is already presented in you solved task. What  else is unclear ?